# Class 11 | Chapter 2 (Structure of Atom) NCERT Chemistry Solutions

Question 1

(i) Calculate the number of electrons which will together weigh one gram.

Mass of an electron = 9.1 × 10-23 g

⇒ 1 g = 1 / 9.1 × 10-23 = 0.1098 × 1023 = 1.098 × 1027 electrons

Thus 1 g of mass is 1.098 × 1027 electrons

(ii) Calculate the mass and charge of one mole of electrons.

One mole of electrons = 6.022 × 1023 electrons
Mass of 1 electron = 9.1 × 10-23 g

⇒ Mass of one mole of electrons = 9.1 × 10-23 × 6.022 × 1023
= 9.1 × 6.022 × 10-23 × 1023
= 54.8002 g

Thus mass of one mole of electrons is 54.8002 g

Similarly charge on one mole of electrons can be calculated.

One mole of electrons = 6.022 × 1023 electrons

Charge on one electron = 1.602 × 10-19 C

⇒ Charge on one mole of electrons = 6.022 × 1023 × 1.602 × 10-19 C
= 6.022 × 1.602 × 10-19 × 1023
= 9.647244 × 104
≈ 9.65 × 104 C

Thus total charge on one mole of electrons is 9.65 × 104 C

Question 2

(i) Calculate the total number of electrons present in one mole of methane

1 mol of Methane(CH4) contains 6.022 × 1023 molecules

1 molecule of Methane(CH4) have 6 + 4 = 10 electrons

⇒ 1 mol of Methane(CH4) contains 6.022 × 1023 × 10 electrons
= 6.022 × 1024 electrons

Thus 1 mol of Methane(CH4) contains 6.022 × 1024 electrons

(ii) Find
(a) The total number and
(b) The total mass of neutrons in 7 mg of 14C
(Assume that mass of a neutron = 1.675 × 10-27 kg)

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed?

Question 3

How many neutrons and protons are there in the following nuclei?

Question 4

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

i) Z = 17, A = 35

ii) Z = 92, A = 233

iii) Z = 4, A = 9

Question 5

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber of yellow light.

Let’s first calculate frequency (ν) of yellow light

Formula for calculating frequency is
ν = c/λ

Where
ν = Frequency
c = Speed of light
λ = Wavelength

Replacing
c = 3 × 108 m/s
λ = 580 nm = 580 × 10-9 m

ν = c/λ = (3 × 108)/(580 × 10-9)

ν = 5.17 × 1014 s-1

Therefore if wavelength of light emitted from a sodium lamp is 580 nm then frequency of light will be 5.17 × 1014 s-1

Let’s now calculate wavenumber of yellow light
Wavenumber is just inverse of wavelength

Therefore
Wavenumber = 1/λ = 1/(580 × 10-9) = 1.7 × 106 m-1

Therefore if wavelength of light emitted from a sodium lamp is 580 nm then it’s wavenumber will be 1.7 × 106 m-1.

Question 6

Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10-10 s

Frequency of light is defined as inverse of it’s period
Frequency = 1/period = 1/(2.0 × 10-10) = 5 × 109 s-1

Therefore if period of light is 2.0 × 10-10 s then it’s frequency will be 5 × 109 s-1.

Wavelength of light is defined as ratio of its speed and frequency.
Wavelength = c/Frequency

Wavelength = (3 × 108)/(5 × 109) = 0.6 × 10-1 = 6.0 × 10-2 m

Therefore if period of light is 2.0 × 10-10 s then its wavelength will be 6.0 × 10-2 m

Wavenumber of light is defined as just inverse of wavelength
Wavenumber = 1/wavelength
= 1/6.0 × 10-2

= 16.66 m-1

Therefore if period of light is 2.0 × 10-10 s then its wavenumber will be 16.66 m-1

Question 7

Calculate the wavelength, frequency and wavenumber of a light wave whose time period is 2.0 × 10-10 s.

Frequency of a light wave (v) = 1/ Time Period = 1/2.0 × 10-10 = 5 × 109 s-1

Wavelength of light wave (λ) = c/v
Where
c = Speed of light
v = Frequency of light wave

Replacing
c = 3 × 108 m/s
v = 5 × 109 s-1

Wavelength of light wave (λ) = c/v = 3 × 108 / 5 × 109 = 6.0 × 10-2 m

Therefore Wavelength of light wave (λ) = 6.0 × 10-2 m

Wavenumber of light wave = 1/λ = 1 / 6.0 × 10-2 = 16.6 m-1

Therefore it time period of a light wave is 2.0 × 10-10 s then it’s wavelength, frequency and wavenumber will be
6.0 × 10-2 m, 5 × 109 s-1 and 16.6 m-1 respectively.

Question 8

What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Energy (E) of a photon = hν

Energy (En) of ‘n’ photons = nhν

Upon dividing both of these equations we get

\begin{equation} E_{n} = nhv \\ \text{ } \\ \text{Rewriting this equation to find out value of n} \\ \text{ } \\ n = \frac{E_{n}}{hv} \end{equation}

In this equation here
n = Number of photons
h = Plank’s Constant
v = Frequency of photon

Relationship between Frequency, Speed of light and Wavelength is
Frequency = Speed/Wavelength

v = c/λ

Therefore formula for n = En/hv can be written as n = Enλ/hc

In this formula plugging in values as
En = 1 J
λ = 4000 pm = 4000 × 10-12 = 4 × 10-9 m
c = 3 × 108 m/s
h = 6.626 × 10-34 Js

\begin{equation} n = \frac {1 × 4 × 10^{-9}} {6.626 × 10^{-34} × 3 × 10^8} \end{equation}

Upon simplification of this mathematical expression we get
n = 2.012 × 1016

Therefore around 2.012 × 1016 number of photos of light with wavelength 4000 pm will provide 1 J of energy.

Question 9

A photon of wavelength 4 x 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV.

Calculate
(i) the energy of the photon (eV)

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV = 1.6020 x 10-19 J)

Question 10

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.

Ionisation energy of one sodium atom = hc/λ

Ionisation energy of one mole Na atoms = NAhc/λ

Replacing
NA = 6.022 × 1023 mol-1
h = 6.626 × 10-34 Js
c = 3 × 108 m/s
λ = 242 nm = 242 × 10-9 m

\begin{equation} \text{Ionisation energy of one mole Na atoms} = \frac{N_{A}hc}{λ} \\ \text{ } \\ \frac{6.022 × 10^{23} × 6.626 × 10^{-34} × 3 × 10^{8}}{242 × 10^{- 9}} = 494.65 \text{ }kJ mole^{- 1} \end{equation}

Therefore ionisation energy of sodium in kJ mol-1 is 494.65

Question 11

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta or photon per second.

Energy of one photon (E) = hc/λ

\begin{equation} E = \frac{hc}{λ} \end{equation}

Replacing
h = 6.626 × 10-34 Js
c = 3 × 108 m/s
λ = 0.57 μm = 0.57 × 10-6 m

\begin{equation} E = \frac{6.626 × 10^{-34} × 3 × 10^{8}}{0.57 × 10^{-6}} \\ \text{ } \\ E = 3.487 × 10^{- 19} J \end{equation}

25 watt bulb means its emitting 25 J of energy per second

1 photon = 3.487 × 10-19 J

Therefore number of photons being emitted per second from bulb
= 25/3.487 × 10-19 = 7.17 × 1019 photons per second

So if a 25 watt bulb is emitting monochromatic yellow light of wavelength 0.57 μm then it’s emitting 7.17 × 1019 photons per second.

Question 12

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency 𝑣0 and work function 𝑊0 of the metal.

Threshold wavelength as given in the given (λ0) = 6800 Å = 6800 × 10-10 m

Threshold frequency of metal (v0) = c/λ0 = (3 × 108)/(6800 × 10-10) = 4.41 × 1014 s-1

Therefore threshold frequency of given metal is 4.41 × 1014 s-1

Work function can be calculated using formula 𝑊0 = hv0
Where
h = Planck’s Constant
v0 = Threshold frequency

v0 = 4.41 × 1014 s-1
h = 6.626 × 10-34 Js

𝑊0 = 4.41 × 1014 × 6.626 × 10-34 = 29.22 × 10-20 J = 2.922 × 10-19 J

Therefore if electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å then threshold frequency of metal is 4.41 × 1014 s-1 and its work function is 2.922 × 10-19 J.

Question 13

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

\begin{equation} \frac{1}{λ} = R\left( \frac{1}{n_{1}^{2}} – \frac{1}{n_{2}^{2}} \right) \\ \text{ } \\ \text{Replacing} \\ \text{ } \\ R = 109677.57 \text{ } cm^{-1} \\ \text{ } \\ n_1 = 2 \\ \text{ } \\ n_2 = 4 \\ \text{ } \\ \frac{1}{λ} = 109677.57\left( \frac{1}{2^2} – \frac{1}{4^2} \right) = 109677.57\left( \frac{1}{4} – \frac{1}{16} \right) \\ \text{ } \\ \frac{1}{λ} = 109677.57\left( \frac{4 – 1}{16} \right) = 109677.57\left( \frac{3}{16} \right) \\ \text{ } \\ \frac{1}{λ} = 109677.57\left( \frac{3}{16} \right) \\ \text{ } \\ λ = \frac{16}{3 × 109677.57} = 4.862 × 10^{- 5} \text{ } cm \end{equation}

λ = 4.862 × 10-5 cm
Writing this in terms of SI units

λ = 4.862 × 10-5 cm = 4.862 × 10-7 m = 486.2 × 10-9 m = 486.2 nm

λ = 486.2 nm

Therefore if an electron in a hydrogen atom transition from energy level n = 4 to energy level n = 2 then wavelength of electron emitted will be 486.2 nm

Question 14

How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?
Compare your answer with the ionisation enthalpy of H atom (energy required to remove the electron from n = 1)

Ionisation energy is the amount of energy required to remove an electron from neutral gaseous atom i.e. to shift the electron from n = 1 to n = ∞

Energy of an electron in Hydrogen atom is given as
E = – 21.8 × 10-19/n2 J/atom

Where n is orbit of electron

Therefore
Energy required to ionise a H atom = Energy of electron in n = ∞ orbitEnergy of electron in n = 1 orbit

\begin{equation} \text{Energy required to ionise a Hydrogen atom} = \\ \text{ } \\ \text{Energy of electron in n = ∞ orbit} – \text{Energy of electron in n = 5 orbit} \\ \text{ } \\ \frac {- 21.8 × 10^{- 19}} {∞^{2}} – \frac {- 21.8 × 10^{- 19}} {5^{2}} \\ \text{ } \\ = 0 – \frac {- 21.8 × 10^{- 19}} {25} \\ \text{ } \\ = \frac {21.8 × 10^{- 19}} {25} = 0.872 × 10^{- 19} \end{equation}

Therefore energy required to ionise a H atom, if its electron is in n = 5 orbit is 0.872 × 10-19 J/atom

Ionisation Enthalpy of H atom is 21.8 × 10-19 J/atom

Comparing Ionisation Enthalpy and Energy required to H atom if electron is in n = 5 orbit.
= 21.8 × 10-19 J/atom divided by 0.872 × 10-19 J/atom

= 25

Therefore Ionisation Enthalpy of H atom is almost 25 times, energy required to ionise a H atom if its electron is in orbit n = 5.

Question 15

What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Maximum Number of emission lines when the excited electron of a H atom in n = 6 drops to ground state is
(n2 – n1)(n2 – n1 + 1)/2
Where n2 , n1 are Orbit Numbers and n2 > n1

As per question its given that electron is n = 6 orbit and is dropping to ground state which is n = 1
Therefore
n2 = 6
n1 = 1

Putting this values in above formula and then simplifying
(6 – 1)(6 – 1 + 1)/2 = (5)(6)/2 = 5 × 3 = 15

Therefore if an excited electron of H atom in n = 6 orbit drops to ground state then there will be 15 emission lines.

Question 16

(i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10-18 J atom-1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Formula to calculate energy of an electron in a hydrogen atom orbit is
En = -2.18 × 10-18/n2 J atom-1
Where
n = Orbit of electron

Thus energy of electron in fifth orbit will be
E5 = – 2.18 × 10-18/52 = – 2.18 × 10-18/25 = 0.0872 × 10-18 = 8.72 × 10-20 J/atom

Therefore energy of electron in fifth orbit of hydrogen atom is 8.72 × 10-20 J/atom

Formula to calculate radius of any orbit of hydrogen atom is
rn = 0.0529/n2 nm

Replacing n = 5 in this formula
rn = 0.0529/25 = 0.002116 nm

Therefore radius of Bohr’s fifth orbit of hydrogen atom is 0.002116 nanometre

Question 17

Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Wavenumber can be calculated using Rydberg formula

\begin{equation} \text{Wavenumber} = R \left ( \frac{1}{n_1^{2}} – \frac{1}{n_2^{2}} \right) \\ \text{ } \\ \text{Where } R = 1.097 × 10^{7} m^{-1} \\ \text{ } \\ n_{1}, \text{ } n_{2} \text{ are number of orbits and } n_{2} > n_{1} \end{equation}

For Balmer Series, n1 = 2
From the Rydberg Formula, its clear that wave number is inversely proportional to wavelength. Hence for wavelength to be longest, wave number have to be smallest. Therefore for wave number to be smallest, 1/n22 have to maximum so n2 have to be minimum.

So possible value of n2 will be 3 because n1 = 2 and we need to have n2 > n1

Therefore
n1 = 2
n2 = 3

\begin{equation} Wavenumber = R \left( \frac{1}{2^2} – \frac{1}{3^2} \right) \\ \text{ } \\ Wavenumber = R \left( \frac{1}{4} – \frac{1}{9} \right) \\ \text{ } \\ Wavenumber = R \left( \frac{9 – 4}{36} \right) = \frac{5R}{36} \\ \text{ } \\ Replacing \text{ } R = 1.097 × 10^{7} m^{- 1} \\ \text{ } \\ Wavenumber = \frac{5 × 1.097 × 10^{7}}{36} \\ \text{ } \\ Wavenumber = 0.152 × 10^{7} = 1.52 × 10^{6} m^{- 1} \\ \text{ } \\ Wavenumber = 1.52 × 10^{6} m^{- 1} \\ \text{ } \\ Wavenumber = \frac{1}{Wave Length} \\ \text{ } \\ \frac{1}{Wave Length} = 1.52 × 10^{6} m^{- 1} \\ \text{ } \\ Wave Length = \frac{1}{1.52 × 10^{6}} = 0.657 × 10^{- 6} \end{equation}

Wavelength = 0.657 × 10-6 m
This number can also be written as 657 nm

Therefore the longest wavelength transition in the Balmer series of atomic hydrogen is 657 nm.

Question 18

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 x 10-11 ergs.

Question 19

The electron energy in hydrogen atom is given by En = (-2.18 × 10-18)/n2 J.

Calculate the energy required to remove an electron completely from the n = 2 orbit.

What is the longest wavelength of light in cm that can be used to cause this transition?

Energy required to completely remove an electron from n = 2 orbit
= Energy of electron at n = ∞ – Energy of electron at n = 2

In the question, it’s given that En = (-2.18 × 10-18)/n2 J

Therefore
E = (-2.18 × 10-18)/∞2 = 0

E2 = (-2.18 × 10-18)/22 = – 0.545 × 10-18 J

Putting these values in above formula
Energy required to completely remove an electron from n = 2 orbit
= 0 – (- 0.545 × 10-18)
= 0.545 × 10-18 J

Therefore if electron energy in hydrogen atom is given by En = (-2.18 × 10-18)/n2 J then energy required to remove an electron completely from n = 2 orbit is 0.545 × 10-18 J

We know that E = hc/λ
Therefore
λ = hc/E

Replacing
h = Plank’s Constant = 6.626 × 10-34 Js
c = Speed of light = 3 × 108 m/s
E = 0.545 × 10-18 J

\begin{equation} λ = \frac{hc}{E} \\ \text{ } \\ λ = \frac {6.626 × 10^{- 34} × 3 × 10^8} {0.545 × 10^{- 18}} = 36.473 × 10^{- 8} \\ \text{ } \\ λ = 36.473 × 10^{- 8} m \\ \text{ } \\ \text{According to question we need to convert this to centimetre, therefore} \\ \text{ } \\ λ = 36.473 × 10^{- 10} cm \end{equation}
\begin{equation} \end{equation}

λ = 36.473 × 10-10 cm is the longest wavelength of light that can be used to remove an electron from n = 2 orbit of hydrogen atom if energy of an electron in nth hydrogen atom orbit is given as En = (-2.18 × 10-18)/n2 J.

Question 20

Calculate the wavelength of an electron moving with a velocity of 2.05 x 107 m s-1

As per De Broglie’s Equation
λ = h/mv
Where
λ = Wavelength of moving particle
h = Planck’s Constant
m = mass of particle
v = velocity of particle

Replacing values in formula λ = h/mv
h = Plank’s Constant = 6.626 × 10-34 Js

m = mass of electron = 9.01 × 10-31 kg

v = 2.05 × 107 m s-1 (Given in question)

\begin{equation} λ = \frac{h}{mv} \\ \text{ } \\ λ = \frac {6.626 × 10^{-34}} {9.01 × 10^{-31} × 2.05 × 10^{7}} \\ \text{ } \\ λ = 0.3587 × 10^{- 10} \\ \text{ } \\ λ = 3.587 × 10^{- 11} m \end{equation}

Therefore if an electron is moving with velocity 2.05 × 107 m s-1 then its wavelength will be 3.587 × 10-11 m

Question 21

The mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × 10-25 J, calculate its wavelength.

Formula for Kinetic Energy is K = mv2/2
Where
m = mass of particle
v = velocity of particle

In this question its given that
m = 9.1 × 10-31 kg
K = 3.0 × 10-25 J

Putting these values in Kinetic Energy formula

\begin{equation} 3.0 × 10^{- 25} = \frac {9.1 × 10^{- 31} × v^{2}} {2} \\ \text{ } \\ 3.0 × 10^{- 25} × 2 = 9.1 × 10^{- 31} × v^{2} \\ \text{ } \\ \frac {3.0 × 10^{- 25} × 2} {9.1 × 10^{- 31}} = v^{2} \\ \text{ } \\ 0.6593 × 10^{6} = v^{2} \\ \text{ } \\ \sqrt{0.6593 × 10^{6}} = v \\ \text{ } \\ 0.8119 × 10^{3} = v \\ \text{ } \\ 811.9 = v \end{equation}

Therefore if mass of an electron is 9.1 × 10-31 kg and its kinetic energy is 3.0 × 10-25 J then its velocity is 812 m/s

As per De Broglie’s Equation
λ = h/mv
Where
λ = Wavelength of moving particle
h = Planck’s Constant
m = mass of particle
v = velocity of particle

Replacing values in formula λ = h/mv
h = Plank’s Constant = 6.626 × 10-34 Js

m = mass of electron = 9.1 × 10-31 kg

v = 812 m s-1 (calculated above)

\begin{equation} λ = \frac{h}{mv} \\ \text{ } \\ λ = \frac {6.626 × 10^{- 34}} {9.1 × 10^{- 31} × 812} \\ \text{ } \\ λ = \frac{6.626 × 10^{- 34}} {7389.2 × 10^{- 31}} \\ \text{ } \\ λ = 8.967 × 10^{-9} m \end{equation}

Therefore wavelength = 8.967 × 10-9 m
Which can also be written as 8967 Å

Therefore if mass of an electron is 9.1 × 10-31 kg and its K.E. is 3.0 × 10-25 J then it’s wavelength will be 8967 Å

Question 22

Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+, K+, Mg2+, Ca2+, S2-, Ar

Isoelectronic Species are those atoms/molecules/ions which have same number of electrons.

Na+ and Mg2+ both have 10 electrons therefore these two are Isoelectronic species

K+, Ca2+, S2- and Ar each have 18 electrons therefore these are Isoelectronic species

Question 23

(i) Write the electronic configurations of the following ions:
(a) H
1s2

(b) Na+
1s2 2s2 2p6

(c) O2-
1s2 2s2 2p6

(d) F
1s2 2s2 2p6

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1
If outermost electrons in an atom are in 3s1
Then its full electronic configuration is 1s2 2s2 2p6 3s1

Calculating total electrons = 2 + 2 + 6 + 1 = 11

Total number of electrons = Atomic Number = 11

(b) 2p3
If outermost electrons in an atom are in 2p3
Then its full electronic configuration is 1s2 2s2 2p3

Calculating total electrons = 2 + 2 + 3 = 7

Total number of electrons = Atomic Number = 7

(c) 3p5
If outermost electrons in an atom are in 3p5
Then its full electronic configuration is 1s2 2s2 2p6 3s2 3p5

Calculating total electrons = 2 + 2 + 6 + 2 + 5 = 17

Total number of electrons = Atomic Number = 17

(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1
[He] means 1s2

Therefore full electronic configuration is 1s2 2s1

Total number of electrons = 2 + 1 = 3

Total Number of electrons = Atomic Number = 3

Atomic Number = 3 means it’s Lithium (Li)

(b) [Ne] 3s23p3
[Ne] means 1s2 2s2 2p6

Therefore full electronic configuration is 1s2 2s2 2p6 3s23p3

Total number of electrons = 2 + 2 + 6 + 2 + 3 = 15

Total Number of electrons = Atomic Number = 15

Atomic Number = 15 means it’s Phosphorus (P)

(c) [Ar] 4s23d1
[Ar] means 1s2 2s2 2p6 3s2 3p6

Therefore full electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s23d1

Total number of electrons = 2 + 2 + 6 + 2 + 6 + 2 + 1 = 21

Total Number of electrons = Atomic Number = 21

Atomic Number = 21 means it’s Scandium (Sc)

Question 24

What is the lowest value of n that allows g-orbitals to exist?

For g-orbital Angular Momentum Quantum Number (I) = 4

We know that relationship between Principle Quantum Number (n) and Angular Momentum Quantum Number (I) is

For any value of n, possible values of Angular Momentum Quantum Number (I) are 0 to n – 1

Therefore
if I = 4

I = 0 to n – 1

4 = 0 to n – 1

Which means n – 1 = 4
Therefore n = 5

Thus lowest value of n which allows g-orbitals to exist is 5.

Question 25

An electron is in one of the 3d orbitals. Give the possible values of n, I and m, for this electron.

For 3d orbitals
n = 3
I = 2
m = – 2 to + 2 = – 2, -1, 0, +1, + 2

Question 26

An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i) the number of protons
Number of protons = Number of electrons

Therefore
Number of protons = 29

(ii) the electronic configuration of the element
Element’s atom have 29 electrons

Therefore its electronic configuration will be
1s2 2s2 2p6 3s2 3p6 3d10 4s1

Question 27

Give the number of electrons in the species H+2, H2 and 𝑂+2

Hydrogen atom have just 1 electron

As H2 is combination of two Hydrogen atoms therefore it will have 2 electrons

But H2+ means H2 have lost 1 electron therefore H2+ will have just 1 electron.

As H atom have just 1 electron, therefore H2 will have 2 electrons.

Oxygen atom have 8 electrons

As O2 is combination of two oxygen atoms therefore it will have 16 electrons

But O2+ means O2 have lost 1 electron therefore O2+ will have 15 electrons.

Question 28

(i) An atomic orbital has n = 3. What are the possible values of l and m?

l = 0 to n – 1 = 0 to 3 -1 = 0 to 2

l = 0, 1, 2

m = – l to + l

If l = 0
m = – 0 to + 0 = 0

If l = 1
m = – 1 to + 1 = – 1, 0, + 1

If l = 2
m = – 2 to + 2 = – 2, – 1, 0, + 1, + 2

Therefore if n = 3 then possible values of l are 0, 1, 2 and m are – 2, – 1, 0, + 1, + 2

(ii) List the quantum numbers (m, and l) of electrons for 3d orbital.
For 3d orbital

n = 3

l = 0 to n – 1 = 0 to 3 – 1 = 0 to 2

l = 0 to 2 = 0, 1, 2

l = 0, 1, 2

If l = 0
m = – l to + l = – 0 to + 0 = 0

If l = 1
m = – 1 to + 1 = – 1, 0, + 1

If l = 2
m = – 2 to + 2 = – 2, – 1, 0, + 1, + 2

Therefore for electrons in 3d orbital
m = – 2, – 1, 0, + 1, + 2
l = 0, 1, 2

(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f

For any particular value of n, allowed values of l are 0 to n – 1 therefore 2s, 2p are only possible orbitals out of 1p, 2s, 2p and 3f.

Question 29

Using s, p, d notations, describe the orbital with the following quantum numbers,
(a) n = 1, l= 0
1s

(b) n = 3, l = 1
3p

(c) n = 4, l = 2
4d

(d) n = 4, l = 3
4f

Question 30

Explain, giving reasons, which of the following sets of quantum numbers are not possible,
(a) n = 0, l = 0, m1 = 0, ms = + 1/2
Not possible because n cannot be zero

(b) n = 1, l = 0, m1 = 0, ms = -1/2
Possible

(c) n= 1, l = 1, m1 = 0, ms = + 1/2
Not possible because when n = 1
l have to be zero and l = 1 not possible

(d) n = 2, l = 1, m1 = 0, ms = -1/2
Possible

(e) n = 3, l = 3, m1 = -3, ms = +1/2
Not possible because n = 3, l = 0,1,2 but not 3

(f) n = 3, l = 1, m1 = 0, ms = + 1/2
Possible

Question 31

How many electrons in an atom may have the following quantum number?
(a) n = 4, ms = -1/2
Total number of electrons in n is 2n2 = 2 (2)2 = 32
But only half of these electrons will have ms = – 1/2 and other half will have ms = + 1/2

Therefore number of electrons in an atom which have n = 4, ms = -1/2 are 16 electrons

(b) n = 3, l = 0
n = 3 and l = 0 is 3s subshell
Which can only contain 2 electrons.

Question 32

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

According to Bohr’s Model of Hydrogen Atom
mvr = nh/2𝝅 (Equation 1)

According to de Broglie equation
λ = h/mv

mv = h/λ (Equation 2)

Replacing value of mv from Equation 2 into Equation 1
hr/λ = nh/2𝝅

Simplifying this
2𝝅r =

2𝝅r means circumference of Bohr orbit (r)

Therefore
Circumference of Bohr Orbit (r) = n times Wavelength of electron in nth orbit.

Question 33

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

Question 34

Calculate the energy required for the process

He+(g) → He2+(g) + e

The ionisation energy of the H atom in the ground state is 2.18 × 10-18 J/atom

Ionisation energy means energy required to completely remove an electron from an orbit of a atom.

\begin{equation} \text{Ionisation Energy} = E_{∞} – E_{1} \\ \text{ } \\ E_{n} = \frac {- 2.18 × 10^{- 18} × Z^{2}} {n^2} \\ \text{ } \\ \text{Ionisation Energy} = \frac{- 2.18 × 10^{- 18} × Z^{2}} {∞^2} – \left ( \frac{- 2.18 × 10^{- 18} × Z^{2}} {1^2} \right) \\ \text{ } \\ \text{Ionisation Energy} = 0 + \left( 2.18 × 10^{- 18} × Z^{2} \right) \\ \text{ } \\ \text{Z = 2 for Helium Atom} \\ \text{ } \\ \text{Ionisation Energy} = 2.18 × 10^{- 18} × 2^{2} = 2.18 × 10^{- 18} × 4 \\ \text{ } \\ \text{Ionisation Energy} = 8.72 × 10^{- 18} \end{equation}

Therefore energy required for He+(g) → He2+(g) + e is 8.72 × 10-18 J

Question 35

If the diameter of a carbon atom is 0.15 nm.

Calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Diameter of carbon atom = 0.15 nm = 0.15 × 10-9 m = 0.15 × 10-11 cm

Number of carbon atoms which can be placed side by side in a straight line across length of scale 20 cm long
= 20/0.15 × 10-11
= 133 × 1011

= 1.33 × 109 number of carbon atoms

Question 36

2 x 108 atoms of carbon are arranged side by side.

Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm

An atom will cover length equal to its diameter

Diameter of carbon atom = 2.4/2 × 108 = 1.2 × 10-8 cm

Radius of carbon atom = (1.2 × 10-8)/2 = 0.6 × 10-8 cm = 0.6 × 10-10 m

Radius of carbon atom = 0.6 × 10-10 m = 0.06 × 10-9 m = 0.06 nm

Therefore Radius of carbon atom = 0.06 nm

Question 37

The diameter of zinc atom is 2.6 Å.

Calculate
(a) radius of zinc atom in pm and

Diameter of zinc atom = 2.6 Å

Radius of zinc atom = 1.3 Å = 1.3 × 10-10 m

Converting this to pm

Radius of zinc atom = 130 × 10-12 = 130 pm

Radius of zinc atom = 130 pm

(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise

Number of atoms present in length of 1.6 cm = 0.016 m

Diameter of zinc atom = 2.6 Å = 2.6 × 10-10 m

Number of atoms present in length of 1.6 cm
= 0.016/(2.6 × 10-10)

= 0.00615 × 1010

= 6.15 × 107 zinc atoms

Question 38

A certain particle carries 2.5 × 10-16 C of static electric charge.

Calculate the number of electrons present in it

Charge on just one electron = 1.602 × 10-19 C

Therefore
Number of electrons present in a particle carrying 2.5 × 10-16 C of static electric charge

= (2.5 × 10-16)/(1.602 × 10-19)

= 1.5605 × 103 electrons

Question 39

In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays.

If the static electric charge on the oil drop is – 1.282 x 10-18 C, what will be the number of electrons present on it?

Charge on just one electron = – 1.602 × 10-19 C

Number of electrons = (– 1.282 x 10-18)/(- 1.602 × 10-19) = 0.8002 × 10 = 8.002

Which is approximately 8

Therefore if an oil drop have static electric charge of – 1.282 x 10-18 C it means there are 8 electrons in it.

Question 40

In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have
been used to be bombarded by the alpha-particles.

If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Lesser number of alpha particles will be deflected because nucleus of lighter atoms have smaller positive charge on their nuclei.

Question 41

Symbols 7935Br and 79Br can be written whereas symbols 3579Br and 35Br are not acceptable.

General rules for representing an element along with its atomic mass (A) and atomic number (Z) is AZX

For Bromine
Atomic mass (A) = 79
Atomic Number (Z) = 35

Therefore correct symbol for Bromine is 7935Br and which also be written as 79Br

But symbols like 3579Br and 35Br are not acceptable because these show atomic number on upper side rather these should be shown on downside.

Question 42

An element with mass number 81 contains 31.7% more neutrons as compared to protons.

Assign the atomic symbol.

We know that
A = p + n

Where
A = Atomic Mass
p = Number of protons
n = Number of neutrons

In the question it’s given that
A = 81

Because element contains 31.7% more neutrons as compared to protons
Therefore
n = p + 31.7p/100 = p + 0.317p = 1.317p

n = 1.317p

Replacing A = 81 and n = 1.317p in equation A = p + n

81 = p + 1.317p

81 = 2.317p

81/2.317 = p

p = 35 approximately

We know that number of protons in an atom equals to atomic number

Therefore atomic number of element is 35

Atomic Number = 35
Atomic Mass = 81

Thus atomic symbol will be 8135Br

Question 43

An ion with mass number 37 possesses one unit of negative charge.
If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Question 44

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol of this ion.

Question 45

Arrange the following type of radiations in increasing order of frequency :
(a)
radiation from microwave oven
(b) amber light from traffic signal
(d) cosmic rays from outer space and
(e) X-ray

c < b < e < d < a

Question 46

Nitrogen laser produces a radiation at a wavelength of 337.1 nm

If the number of photons emitted is 5.6 x 1024, calculate the power of this laser

Wavelength of nitrogen = 337.1 nm = 337.1 × 10-9 m

Number of photons(n) = 5.6 × 1024

Energy of photons = nhc/λ

Replacing
n = 5.6 × 1024
h = 6.626 × 10-34 Js
c = 3 × 108 m/s
λ = 337.1 × 10-9 m

\begin{equation} \text{Energy of photons} = \frac {5.6 × 10^{24} × 6.626 × 10^{- 34} × 3 × 10^{8}} {337.1 × 10^{- 9}} \\ \text{ } \\ \end{equation}

Simplifying this large calculation

Energy of photons = 3.33 × 106 J

Therefore power of this laser = 3.33 × 106 Watt

Question 47

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate

(a) the frequency of emission

Wavelength of emission = 616 nm = 616 × 10-9 m

Frequency of emission (v) = c/λ

Replacing
c = 3 × 108 m/s
λ = 616 × 10-9 m

Therefore
Frequency of emission (v) = (3 × 108)/(616 × 10-9) = 0.0048 × 1017 = 4.8 × 1014 s-1

Thus if wavelength of Neon gas is 616 nm then its frequency will be 4.8 × 1014 s-1

(b) distance travelled by this radiation in 30 s

Distance = Speed × Time = 3 × 108 × 30 = 9 × 109 m

Therefore
Distance travelled by Neon gas radiation in 30 seconds is 9 × 109 m

(c) energy of quantum
Energy of quantum (E) = hv

Replacing
h = 6.626 × 10-34 Js
v = 4.8 × 1014 s-1 (Calculated Above)

E = 6.626 × 10-34 × 4.8 × 1014 = 31.8 × 10-20 = 3.18 × 10-19 J

Therefore
If a Neon gas sign board is emitting light at wavelength 616 nm then energy of quantum emitted will be 3.18 × 10-19 J

(d) number of quanta present if it produces 2 J of energy

Number of quanta in 2J of energy
= 2/3.18 × 10-19

= 0.628 × 1019 = 6.2 × 1018 photons

Question 48

In astronomical observations, signals observed from the distant stars are generally weak.

If the photon detector receives a total of 3.15 x 10-18 J from the radiations of 600 nm.

Calculate the number of photons received by the detector.

Question 49

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range.

If the radiation source has the duration of 2 ns and number of photons emitted during the pulse source is 2.5 × 1015.

Calculate the energy of the source.

Duration = 2 ns = 2 × 10-9 s
Frequency of energy (v) = 1/Duration = 1/2 × 10-9 = 0.5 × 109 s-1

Energy of source = n × hv

Where
n = Number of photons
h = Planck’s Constant
v = Frequency of radiation

Replacing
n = 2.5 × 1015
h = 6.626 × 10-34 Js
v = 0.5 × 109 s-1

Therefore
Energy of source = 2.5 × 1015 × 6.626 × 10-34 × 0.5 × 109
= 8.2825 × 10-10 J

If radiation source has the duration of 2 ns and number of photons emitted during the pulse source is 2.5 × 1015 then energy of source is 8.2825 × 10-10 J

Question 50

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm.

Calculate the frequency of each transition and energy difference between two excited states.

Question 51

The work function for caesium atom is 1.9 eV.

Calculate
(a) the threshold frequency and
Work function (W) = 1.9 eV

1 eV = 1.602 × 10-19 J

Therefore
Work function (W) = 1.9 × 1.602 × 10-19 J = 3.0438 × 10-19 J

Work function = 3.0438 × 10-19 J

Work function = hv
v = Work function/h

Replacing
Work function = 3.0438 × 10-19 J
h = 6.626 × 10-34 Js

Therefore
v = (3.0438 × 10-19)/(6.626 × 10-34)

v = 0.4593 × 1015 = 4.593 × 1014 s-1

Therefore threshold frequency of an atom is 4.593 × 1014 s-1 if it’s work function is 1.9 eV

(b) the threshold wavelength of the radiation
Threshold Wavelength = Speed of light/Threshold frequency

Threshold frequency = 4.593 × 1014 s-1
Speed of light = 3 × 108 m/s

Therefore
Threshold Wavelength = (3 × 108)/(4.593 × 1014) = 0.6531 × 10-6 = 6.531 × 10-5 m

Therefore if an atom have work function 1.9 eV then it’s threshold wavelength is 6.531 × 10-5 m

(c) If the caesium elements is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

When an electron got ejected from an atom, then it’s kinetic energy is equal to it’s energy due to frequency minus threshold energy.

Therefore
Kinetic energy of photoelectron = h(Frequency of electron) – h(Threshold frequency)

= h(Frequency of electron – Threshold frequency)

Kinetic energy of photoelectron = h(Frequency of electron – Threshold frequency)

Just above in part (a) of this question, we’ve calculated threshold frequency which is 4.593 × 1014 s-1

Thus
Threshold frequency = 4.593 × 1014 s-1

Frequency of electron = (Speed of light)/(Wavelength)

Speed of light = 3 × 108 m/s

Wavelength = 500 nm = 500 × 10-9 m

Therefore
Frequency of electron = (3 × 108)/(500 × 10-9) = 0.006 × 1017 = 6 × 1014 s-1

Frequency of electron = 6 × 1014 s-1

Replacing
Frequency of electron = 6 × 1014 s-1
Threshold frequency = 4.593 × 1014 s-1

in formula
Kinetic energy of photoelectron = h(Frequency of electron – Threshold frequency)

Kinetic energy of photoelectron = h(6 × 1014 – 4.593 × 1014)

= h(6 – 4.593) × 1014

= h(1.407 × 1014)

Replacing h = 6.626 × 10-34 Js

= 6.626 × 10-34 × 1.407 × 1014

= 9.322 × 10-20

Therefore
Kinetic energy of photoelectron = 9.322 × 10-20 J

Also we need to figure out what’s velocity of this photoelectron

We know that

Kinetic energy = mv2/2

Where
m = mass of electron
v = velocity of electron

Replacing
Kinetic energy = 9.322 × 10-20 J

m = 9.1 × 10-31 kg

9.322 × 10-20 = 9.1 × 10-31 × v2/2

(9.322 × 10-20 × 2)/(9.1 × 10-31) = v2

2.05 × 1011 = v2

Therefore v = 4.52 × 105 m/s

Thus if the caesium element is irradiated with a wavelength 500 nm and it’s work function is 1.9 eV then kinetic energy of ejected photoelectron will be 9.322 × 10-20 J and it’s velocity will be 4.52 × 105 m/s

Question 52

Following results are observed when sodium metal is irradiated with different wavelengths.

λ(nm)                      500       450      400
v x 105 (m s-1)     2.55       4.35      5.20

Calculate
(a) threshold wavelength and,

(b) Planck’s constant

Question 53

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used.

Calculate the work function for silver metal.

Therefore work function of silver metal is 4.48 eV

Question 54

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 ms-1, calculate the energy with which it is bound to the nucleus.

Therefore energy with which the electron is bound to nucleus is 7.6 × 103 eV

Question 55

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be
represented as 𝜐 = 3.29 × 1015 (𝐻𝑍)[1/32 − 1/𝑛2

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Question 56

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Name the series to which this transition belongs and the region of the spectrum.

Question 57

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material.

If the velocity of the electron in this microscope is 1.6 × 106 m/s, calculate de Broglie wavelength associated with this electron.

Therefore if velocity of the electron in a microscope is 1.6 × 106 m/s then de Broglie wavelength associated with electron is 455 picometre.

Question 58

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules.

If the wavelength used here is 800 pm, calculate the characteristic velocity associated with neutron.

λ = 800 pm = 800 × 10-12 m

According to de Broglie formula
λ = h/mv
Where
λ = Wavelength of particle
h = Planck’s Constant
m = Mass of particle
v = Velocity of particle

Replacing
λ = 800 × 10-12 m

h = 6.626 × 10-34 Js

m = mass of particle (neutron) = 1.67 × 10-27 kg

800 × 10-12 = (6.626 × 10-34)/(1.67 × 10-27) v

v = (6.626 × 10-34)/(1.67 × 10-27) × (800 × 10-12)

v = 0.00495 × 105 = 4.95 × 102

v = 4.95 × 102 m/s

Therefore velocity associated with neutron will be 4.95 × 102 m/s if its wavelength is 800 pm

Question 59

If the velocity of the electron in Bohr’s first orbit is 2.19 x 106 ms-1

Calculate the de Broglie wavelength associated with it.

According to de Broglie formula
λ = h/mv
Where
λ = Wavelength of particle
h = Planck’s Constant
m = Mass of particle
v = Velocity of particle

Replacing
λ = ?

h = 6.626 × 10-34 Js

m = mass of particle (electron) = 9.1 × 10-31 kg

v = 2.19 × 106 ms-1

λ = (6.626 × 10-34)/(9.1 × 10-31) × (2.19 × 106)

λ = 0.3324 × 10-9 = 332.4 × 10-12

λ = 332.4 × 10-12 m = 332.4 picometre

λ = 332.4 picometre

If velocity of electron in Bohr’s first orbit is 2.19 x 106 ms-1 then its de Broglie wavelength is 332.4 picometre

Question 60

The velocity associated with a proton moving in a potential difference of 1000V is 4.37 x 105 ms-1.

If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

According to de Broglie formula
λ = h/mv
Where
λ = Wavelength of particle
h = Planck’s Constant
m = Mass of particle
v = Velocity of particle

Replacing
h = 6.626 × 10-34 Js

m = 0.1 kg

v = 4.37 x 105 ms-1

λ = (6.626 × 10-34)/0.1 × 4.37 x 105 = 15.16 × 10-39 = 1.516 × 10-38 m

λ = 1.516 × 10-38 m

Therefore is a hockey ball of mass 0.1 kg is moving with velocity 4.37 x 105 ms-1 then wavelength associated with it will be 1.516 × 10-38 m

Question 61

If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron.

Suppose the momentum of the electron is ℎ/ 4Π𝑟 × 0.05𝑛𝑚 , is there any problem in defining this value.

Question 62

The quantum numbers of six electrons are given below.

Arrange them in order of increasing energies.

If any of these combination (s) has l have the same energy lists :

1. n = 4, l = 2, m1 = -2, ms = -1/2
2. n = 3, l = 2, m1= 1, ms =+1/2
3. n = 4, l= 1, m1 = 0, ms = +1/2
4. n = 3, l = 2,m1 = -2, ms = -M2
5. n = 3, l = 1, m1=-1, ms =+1/2
6. n = 4, l = 1, m1 = 0, mss =+1/2

Energy of any subshell depends upon sum of it’s Principle Quantum Number (n) and its Angular Momentum Number(l)

Therefore
Energy = n + l

1. n = 4, l = 2, m1 = -2, ms = -1/2 (Energy = 4 + 2 = 6)
2. n = 3, l = 2, m1= 1, ms =+1/2 (Energy = 3 + 2 = 5)
3. n = 4, l= 1, m1 = 0, ms = +1/2 (Energy = 4 + 1 = 5)
4. n = 3, l = 2,m1 = -2, ms = -1/2 (Energy = 3 + 2 = 5)
5. n = 3, l = 1, m1=-1, ms =+1/2 (Energy = 3 + 1 = 4)
6. n = 4, l = 1, m1 = 0, mss =+1/2 (Energy = 4 + 1 = 5)

Therefore order of energies is 5 < 2 = 4 < 6 = 3 < 1

Question 63

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital.

Which of these electron experiences the lowest effective nuclear charge?

Effective nuclear charge for the electron in 4p orbital is lowest as its farthest from nucleus of atom.

Question 64

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?
(i) 2s and 3s,
(ii) 4d and 4f,
(iii) 3d and 3p.

Orbital which is closer to nucleus of atom will experience large effective nuclear charge
Therefore

(i) Out of 2s and 3s orbital, 2s orbital will experience larger effective nuclear charge

(ii) Out of 4d and 4f orbital, 4d orbital will experience larger effective nuclear charge

(iii) Out of 3d and 3p orbital, 3p orbital will experience larger effective nuclear charge

Question 65

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Electronic configuration of Al is 1s2 2s2 2p6 3s2 3p1

Electronic configuration of Si is 1s2 2s2 2p6 3s2 3p2

Unpaired electrons in both of Al and Si are present in 3p but in case of Si nucleus have 14 protons while Al just have 13 protons. Therefore total effective nuclear charge experienced by unpaired electrons in Si will be higher as compared to that on unpaired electrons in Al.

Question 66

Indicate the number of unpaired electrons in
(a) P,
(b) Si,
(c) Cr,
(d) Fe and
(e) Kr

Question 67

(a) How many sub-shells are associated with n = 4?
Number of sub-shells in a shell which have Principle Quantum Number (n) = 0 to n – 1

For n = 4
= 0 to 4 -1 = 0 to 3 = 0, 1, 2, 3

Therefore for n = 4, shell have 4 sub-shells

(b) How many electrons will be present in the sub-shells having m value of -1/2 for n = 4?
Total number of electrons in n = 4 is 2n2 = 2(4)2 = 2(16) = 32

Out of this 32 electrons, half of these have m = – 1/2

Thus
There’re 32 electrons present in the sub-shells having m value of -1/2 for n = 4

## Some Concepts Related to Class 11 Chemistry Chapter 2 (Structure of Atom)

What is threshold wavelength?
When the wavelength of the light hitting the metal surface is less than its threshold wavelength, photo electrons are emitted from the surface. A photoelectric effect can only occur if incoming radiation has a certain minimum wavelength, this minimum wavelength is called threshold wavelength.

What is threshold frequency?
The term “threshold frequency” refers to the lowest feasible frequency, below which there is no chance of photoelectric emission occurring, irrespective of the intensity of incoming radiation.

What is work function?
The work function is defined as the minimal amount of thermodynamic work (also known as energy) that is necessary to remove an electron from a solid to a point in the vacuum that is immediately outside the surface of the solid.

What is Rydberg Formula?
Wavelength of light emitted by an electron moving between different energy levels (called orbits) of an atom is determined by the Rydberg formula.

The energy of an electron varies as it moves from one atomic orbit to another. A photon of light is emitted when an electron moves from an orbit with high energy to one with lower energy. Reverse of this is also true, which means if an atom absorbs photons of light then electrons from lower orbit transition to upper orbit.

In order to measure wavelength of light emitted as electron transition from higher to lower energy level, Rydberg purposed a formula which is.

1/λ = R(1/n12 – 1/n22)
Where
λ = Wavelength of light emitted by an electron
n1 and n2 are Integers (n2 > n1) both n1,n2 represent Principle Quantum Numbers of orbits
R = Rydberg Constant = 109677.57 cm-1

What is Ionisation Enthalpy?
The Ionization Enthalpy is a metric that measures an element’s tendency to lose an electron from its valence shell. Mathematically, it is the amount of energy required to remove one electron from a gaseous atom’s valence shell.

For example – Ionization Enthalpy of Hydrogen is energy required to remove an electron from a hydrogen atom which is in gaseous state.
Therefore lower the Ionisation Enthalpy of an atom, easier it is to remove an electron from its outermost shell.

Ionisation Enthalpy usually refers to energy required to remove first valence electron, but there could be more than one valence electrons in an atom. So if it’s energy required to remove second or third valence electron from an atom. Then its called Second or Third ionisation enthalpy.

Formula to calculate energy of an electron in a hydrogen atom orbit is
En = -2.18 × 10-18/n2 J atom-1
Where
n = Orbit of electron

Formula to calculate radius of any orbit of hydrogen atom is
rn = 0.0529/n2 nm
Where
n = Orbit of electron

What is de Broglie Equation?
According to the de Broglie equation, matter may behave as waves, just like light and radiation. A beam of electrons, like a beam of light, may be diffracted in the same way.

As a particle and as a wave, electromagnetic radiation may exist in two distinct states at the same time (expressed as frequency, wavelength). This dual nature characteristic was also found in electrons.

There is a wave-like aspect to each moving particle, whether microscopic or macro-scale, according to Louis de Broglie thesis. ‘Matter Waves’ was the name of this phenomenon. Therefore, a particle’s velocity and momentum may be linked to its wavelength if it were to act like a wave.

Mathematically, de Broglie equation can be expressed as
λ = h/mv
Where
λ = Wavelength of particle wave
h = Planck’s Constant
m = mass of particle
v = velocity of particle

What are Isoelectric Species?
Isoelectronic Species are those atoms/molecules/ions which have same number of electrons.