**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 × 10^{23} = 1.098 × 10^{27} electrons

Thus 1 g of mass is **1.098 × 10 ^{27} electrons**

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

One mole of electrons = 6.022 × 10^{23} electrons

Mass of 1 electron = 9.1 × 10^{-23} g

⇒ Mass of one mole of electrons = 9.1 × 10^{-23} × 6.022 × 10^{23}

= 9.1 × 6.022 × 10^{-23} × 10^{23}

= 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 × 10^{23} electrons

Charge on one electron = 1.602 × 10^{-19} C

⇒ Charge on one mole of electrons = 6.022 × 10^{23} × 1.602 × 10^{-19} C

= 6.022 × 1.602 × 10^{-19} × 10^{23}

= 9.647244 × 10^{4}

≈ 9.65 × 10^{4} C

Thus total charge on one mole of electrons is **9.65 × 10 ^{4} C**

**Question 2**

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

1 mol of Methane(CH_{4}) contains 6.022 × 10^{23} molecules

1 molecule of Methane(CH_{4}) have 6 + 4 = 10 electrons

⇒ 1 mol of Methane(CH_{4}) contains 6.022 × 10^{23} × 10 electrons

= 6.022 × 10^{24} electrons

Thus 1 mol of Methane(CH_{4}) contains **6.022 × 10 ^{24} electrons**

(ii) Find

(a) The total number and

(b) The total mass of neutrons in 7 mg of ^{14}C

(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 NH_{3} 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 × 10^{8} m/s

λ = 580 nm = 580 × 10^{-9} m

ν = c/λ = (3 × 10^{8})/(580 × 10^{-9})**ν = 5.17 × 10 ^{14} s^{-1}**

Therefore if wavelength of light emitted from a sodium lamp is 580 nm then frequency of light will be

**5.17 × 10**

^{14}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 × 10 ^{6} m^{-1}**

Therefore if wavelength of light emitted from a sodium lamp is 580 nm then it’s wavenumber will be

**1.7 × 10**.

^{6}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 × 10**

^{9}s^{-1}Therefore if period of light is

**2.0 × 10**then it’s frequency will be

^{-10}s**5 × 10**.

^{9}s^{-1}Wavelength of light is defined as ratio of its speed and frequency.

Wavelength = c/Frequency

Wavelength = (3 × 10

^{8})/(5 × 10

^{9}) = 0.6 × 10

^{-1}=

**6.0 × 10**

^{-2}mTherefore if period of light is

**2.0 × 10**then its wavelength will be

^{-10}s**6.0 × 10**

^{-2}mWavenumber 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**then its wavenumber will be

^{-10}s**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 × 10 ^{9} s^{-1}**

Wavelength of light wave (λ) = c/v

Where

c = Speed of light

v = Frequency of light wave

Replacing

c = 3 × 10

^{8}m/s

v = 5 × 10

^{9}s

^{-1}

Wavelength of light wave (λ) = c/v = 3 × 10

^{8}/ 5 × 10

^{9}= 6.0 × 10

^{-2}m

Therefore Wavelength of light wave (λ) =

**6.0 × 10**

^{-2}mWavenumber 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 × 10**and

^{9}s^{-1}**16.6 m**respectively.

^{-1}**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 (E_{n}) of ‘n’ photons = nhν

Upon dividing both of these equations we get

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 = E _{n}λ/hc**

In this formula plugging in values as

E

_{n}= 1 J

λ = 4000 pm = 4000 × 10

^{-12}= 4 × 10

^{-9}m

c = 3 × 10

^{8}m/s

h = 6.626 × 10

^{-34}Js

Upon simplification of this mathematical expression we get

n = 2.012 × 10^{16}

Therefore around 2.012 × 10^{16} 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 = N_{A}hc/λ

Replacing

N_{A} = 6.022 × 10^{23} mol^{-1}

h = 6.626 × 10^{-34} Js

c = 3 × 10^{8} m/s

λ = 242 nm = 242 × 10^{-9} m

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

Replacing

h = 6.626 × 10^{-34} Js

c = 3 × 10^{8} m/s

λ = 0.57 μm = 0.57 × 10^{-6} m

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 × 10 ^{19} 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 × 10**.

^{19}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 𝑣_{0} and work function 𝑊_{0} of the metal.

Threshold wavelength as given in the given (λ_{0}) = 6800 Å = 6800 × 10^{-10} m

Threshold frequency of metal (v_{0}) = c/λ_{0} = (3 × 10^{8})/(6800 × 10^{-10}) = 4.41 × 10^{14} s^{-1}

Therefore threshold frequency of given metal is **4.41 × 10 ^{14} s^{-1}**

Work function can be calculated using formula 𝑊

_{0}= hv

_{0}

Where

h = Planck’s Constant

v

_{0}= Threshold frequency

v

_{0}= 4.41 × 10

^{14}s

^{-1}

h = 6.626 × 10

^{-34}Js

𝑊

_{0}= 4.41 × 10

^{14}× 6.626 × 10

^{-34}= 29.22 × 10

^{-20}J =

**2.922 × 10**

^{-19}JTherefore 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 × 10**and its work function is

^{14}s^{-1}**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?

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}/n^{2} J/atom

Where n is orbit of electron

Therefore

Energy required to ionise a H atom = **Energy of electron in n = ∞ orbit** – **Energy of electron in n = 1 orbit**

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/atomComparing Ionisation Enthalpy and Energy required to H atom if electron is in n = 5 orbit.

=

**21.8 × 10**divided by

^{-19}J/atom**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**(n _{2} – n_{1})(n_{2} – n_{1} + 1)/2**

Where n

_{2}, n

_{1}are Orbit Numbers and n

_{2}> n

_{1}

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

Therefore

n

_{2}= 6

n

_{1}= 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

E_{n} = -2.18 × 10^{-18}/n^{2} J atom^{-1}

Where

n = Orbit of electron

Thus energy of electron in fifth orbit will be

E_{5} = – 2.18 × 10^{-18}/5^{2} = – 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

r

_{n}= 0.0529/n

^{2}nm

Replacing n = 5 in this formula

r

_{n}= 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**

For Balmer Series, n_{1} = 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/n_{2}^{2} have to maximum so n_{2} have to be minimum.

So possible value of n_{2} will be 3 because n_{1} = 2 and we need to have n_{2} > n_{1}

Therefore

n_{1} = 2

n_{2} = 3

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 E_{n} = (-2.18 × 10^{-18})/n^{2} 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 E

_{n}= (-2.18 × 10

^{-18})/n

^{2}J

Therefore

E

_{∞}= (-2.18 × 10

^{-18})/∞

^{2}= 0

E

_{2}= (-2.18 × 10

^{-18})/2

^{2}= – 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 E

_{n}= (-2.18 × 10

^{-18})/n

^{2}J then energy required to remove an electron completely from n = 2 orbit is

**0.545 × 10**

^{-18}JWe know that E = hc/λ

Therefore

λ = hc/E

Replacing

h = Plank’s Constant = 6.626 × 10

^{-34}Js

c = Speed of light = 3 × 10

^{8}m/s

E = 0.545 × 10

^{-18}J

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 E_{n} = (-2.18 × 10^{-18})/n^{2} J.

**Question 20**

Calculate the wavelength of an electron moving with a velocity of 2.05 x 10^{7} 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 × 10^{7} m s^{-1} (Given in question)

Therefore if an electron is moving with velocity 2.05 × 10^{7} 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 = mv ^{2}/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

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)

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^{+}, Mg^{2+}, Ca^{2+}, S^{2-}, Ar

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

Species | Number of Electrons |
---|---|

Na^{+} | 10 |

K^{+} | 18 |

Mg^{2+} | 10 |

Ca^{2+} | 18 |

S^{2-} | 18 |

Ar | 18 |

Na^{+} and Mg^{2+} both have 10 electrons therefore these two are Isoelectronic species

K^{+}, Ca^{2+}, S^{2-} and Ar each have 18 electrons therefore these are Isoelectronic species

**Question 23****(i)** Write the electronic configurations of the following ions:**(a)** H^{–}**1s ^{2}**

**(b)**Na

^{+}

**1s**

^{2}2s^{2}2p^{6}**(c)**O

^{2-}

**1s**

^{2}2s^{2}2p^{6}**(d)**F

^{–}

**1s**

^{2}2s^{2}2p^{6}**(ii)**What are the atomic numbers of elements whose outermost electrons are represented by

**(a)**3s

^{1}

If outermost electrons in an atom are in 3s

^{1}

Then its full electronic configuration is 1s

^{2}2s

^{2}2p

^{6}3s

^{1}

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

Total number of electrons = Atomic Number = 11

**(b)**2p

^{3}

If outermost electrons in an atom are in 2p

^{3}

Then its full electronic configuration is 1s

^{2}2s

^{2}2p

^{3}

Calculating total electrons = 2 + 2 + 3 = 7

Total number of electrons = Atomic Number = 7

**(c)**3p

^{5}

If outermost electrons in an atom are in 3p

^{5}

Then its full electronic configuration is 1s

^{2}2s

^{2}2p

^{6}3s

^{2}3p

^{5}

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] 2s

^{1}

[He] means 1s

^{2}

Therefore full electronic configuration is 1s

^{2}2s

^{1}

Total number of electrons = 2 + 1 = 3

Total Number of electrons = Atomic Number = 3

Atomic Number = 3 means it’s

**Lithium (Li)**

**(b)**[Ne] 3s

^{2}3p

^{3}

[Ne] means 1s

^{2}2s

^{2}2p

^{6}

Therefore full electronic configuration is 1s

^{2}2s

^{2}2p

^{6}3s

^{2}3p

^{3}

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] 4s

^{2}3d

^{1}

[Ar] means 1s

^{2}2s

^{2}2p

^{6}3s

^{2}3p

^{6}

Therefore full electronic configuration is 1s

^{2}2s

^{2}2p

^{6}3s

^{2}3p

^{6}4s

^{2}3d

^{1}

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**1s ^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{1}**

**Question 27**

Give the number of electrons in the species H^{+}_{2}, H_{2} and 𝑂^{+}_{2}

Hydrogen atom have just 1 electron

As H_{2} is combination of two Hydrogen atoms therefore it will have 2 electrons

But H_{2}^{+} means H_{2} have lost 1 electron therefore **H _{2}^{+} will have just 1 electron**.

As H atom have just 1 electron, therefore

**H**.

_{2}will have 2 electronsOxygen atom have 8 electrons

As O

_{2}is combination of two oxygen atoms therefore it will have 16 electrons

But O

_{2}

^{+}means O

_{2}have lost 1 electron therefore

**O**.

_{2}^{+}will have 15 electronsRelated – What is Electron Configuration of Oxygen?

**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, + 2l = 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, m_{1} = 0, m_{s} = + 1/2

Not possible because n cannot be zero**(b)** n = 1, l = 0, m_{1} = 0, m_{s} = -1/2

Possible**(c)** n= 1, l = 1, m_{1} = 0, m_{s} = + 1/2

Not possible because when n = 1

l have to be zero and l = 1 not possible**(d)** n = 2, l = 1, m_{1} = 0, m_{s} = -1/2

Possible**(e)** n = 3, l = 3, m_{1} = -3, m_{s} = +1/2

Not possible because n = 3, l = 0,1,2 but not 3**(f)** n = 3, l = 1, m_{1} = 0, m_{s} = + 1/2

Possible

**Question 31**

How many electrons in an atom may have the following quantum number?**(a)** n = 4, m_{s} = -1/2

Total number of electrons in n is 2n^{2} = 2 (2)^{2} = 32

But only half of these electrons will have m_{s} = – 1/2 and other half will have m_{s} = + 1/2

Therefore number of electrons in an atom which have n = 4, m_{s} = -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 = nλ****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) → He^{2+}(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.

Therefore energy required for **He ^{+}(g) → He^{2+}(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 × 10^{11}

= 1.33 × 10^{9} number of carbon atoms

Related – What is Electronic Configuration of Carbon Atom

**Question 36**

2 x 10^{8} 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 × 10^{8} = 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 × 10

^{10}

=

**6.15 × 10**

^{7}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 × 10**electrons

^{3}**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**it means there are 8 electrons in it.

^{-18}C**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 ^{79}_{35}Br and ^{79}Br can be written whereas symbols ^{35}_{79}Br and ^{35}Br are not acceptable.

Answer briefly

General rules for representing an element along with its atomic mass (A) and atomic number (Z) is ^{A}_{Z}X

For Bromine

Atomic mass (A) = 79

Atomic Number (Z) = 35

Therefore correct symbol for Bromine is ^{79}_{35}Br and which also be written as ^{79}Br

But symbols like ^{35}_{79}Br and ^{35}Br 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 ^{81}_{35}Br

**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

**(c)**radiation from FM radio

**(d)**cosmic rays from outer space and

**(e)**X-ray

**c < b < e < d < a**

Type of Radiation | Frequency |
---|---|

Radiation from microwave oven | 2.45 Ghz |

Amber light from traffic signal | 405 – 530 THz |

Radiation from FM Radio | 88 – 108 MHz |

Cosmic rays from outer space | 10^{18} to 10^{39} Hz |

X – ray | 10^{16} to 10^{19} Hz |

**Question 46**

Nitrogen laser produces a radiation at a wavelength of **337.1 nm**

If the number of photons emitted is **5.6 x 10 ^{24}**, calculate the power of this laser

Wavelength of nitrogen = 337.1 nm = 337.1 × 10

^{-9}m

Number of photons(n) = 5.6 × 10

^{24}

Energy of photons = nhc/λ

Replacing

n = 5.6 × 10

^{24}

h = 6.626 × 10

^{-34}Js

c = 3 × 10

^{8}m/s

λ = 337.1 × 10

^{-9}m

Simplifying this large calculation

Energy of photons = 3.33 × 10^{6} J

Therefore **power of this laser = 3.33 × 10 ^{6} Watt**

**Related** – What is Electronic Configuration of Nitrogen Atom

**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 × 10^{8} m/s

λ = 616 × 10^{-9} m

Therefore

Frequency of emission (v) = (3 × 10^{8})/(616 × 10^{-9}) = 0.0048 × 10^{17} = 4.8 × 10^{14} s^{-1}

Thus if wavelength of Neon gas is 616 nm then its frequency will be **4.8 × 10 ^{14} s^{-1}**

**(b)**distance travelled by this radiation in 30 s

Distance = Speed × Time = 3 × 10

^{8}× 30 = 9 × 10

^{9}m

Therefore

Distance travelled by Neon gas radiation in 30 seconds is

**9 × 10**

^{9}m**(c)**energy of quantum

Energy of quantum (E) = hv

Replacing

h = 6.626 × 10

^{-34}Js

v = 4.8 × 10

^{14}s

^{-1}(Calculated Above)

E = 6.626 × 10

^{-34}× 4.8 × 10

^{14}= 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 × 10

^{19}=

**6.2 × 10**

^{18}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 × 10^{15}.

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 × 10^{9} s^{-1}**Energy of source = n × hv**

Where

n = Number of photons

h = Planck’s Constant

v = Frequency of radiation

Replacing

n = 2.5 × 10^{15}

h = 6.626 × 10^{-34} Js

v = 0.5 × 10^{9} s^{-1}

Therefore

Energy of source = 2.5 × 10^{15} × 6.626 × 10^{-34} × 0.5 × 10^{9}

= 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 × 10^{15} 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 × 10^{15} = 4.593 × 10^{14} s^{-1}

Therefore threshold frequency of an atom is **4.593 × 10 ^{14} 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 × 10

^{14}s

^{-1}

Speed of light = 3 × 10

^{8}m/s

Therefore

Threshold Wavelength = (3 × 10

^{8})/(4.593 × 10

^{14}) = 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 × 10

^{14}s

^{-1}

Thus

**Threshold frequency = 4.593 × 10**

^{14}s^{-1}Frequency of electron = (Speed of light)/(Wavelength)

Speed of light = 3 × 10

^{8}m/s

Wavelength = 500 nm = 500 × 10

^{-9}m

Therefore

Frequency of electron = (3 × 10

^{8})/(500 × 10

^{-9}) = 0.006 × 10

^{17}= 6 × 10

^{14}s

^{-1}

**Frequency of electron = 6 × 10**

^{14}s^{-1}Replacing

**Frequency of electron = 6 × 10**

^{14}s^{-1}**Threshold frequency = 4.593 × 10**

^{14}s^{-1}in formula

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

Kinetic energy of photoelectron = h(6 × 10

^{14}– 4.593 × 10

^{14})

= h(6 – 4.593) × 10

^{14}

= h(1.407 × 10

^{14})

Replacing h = 6.626 × 10

^{-34}Js

= 6.626 × 10

^{-34}× 1.407 × 10

^{14}

= 9.322 × 10

^{-20}

Therefore

**Kinetic energy of photoelectron = 9.322 × 10**

^{-20}JAlso we need to figure out what’s velocity of this photoelectron

We know that

**Kinetic energy = mv**

^{2}/2Where

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}**×**

**v**

^{2}/2(9.322 × 10

^{-20}× 2)/(9.1 × 10

^{-31}) = v

^{2}

2.05 × 10

^{11}= v

^{2}

Therefore

**v = 4.52 × 10**

^{5}m/sThus 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**and it’s velocity will be

^{-20}J**4.52 × 10**

^{5}m/s**Question 52**

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

λ(nm) 500 450 400

v x 10^{5} (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 10^{7} 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 × 10 ^{3} 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 × 10^{15} (𝐻𝑍)[1/3^{2} − 1/𝑛^{2}]

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 × 10^{6} m/s, calculate de Broglie wavelength associated with this electron.

Therefore if velocity of the electron in a microscope is 1.6 × 10^{6} 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 × 10^{5} = 4.95 × 10^{2}**v = 4.95 × 10 ^{2} m/s**

Therefore velocity associated with neutron will be

**4.95 × 10**if its wavelength is 800 pm

^{2}m/s**Question 59**

If the velocity of the electron in Bohr’s first orbit is 2.19 x 10^{6} 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 × 10^{6} ms^{-1}

λ = (6.626 × 10^{-34})/(9.1 × 10^{-31}) × (2.19 × 10^{6})

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 10^{6} 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 10^{5} 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 10^{5} ms^{-1}

λ = (6.626 × 10^{-34})/0.1 × 4.37 x 10^{5} = 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 10

^{5}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, m_{1} = -2, m_{s} = -1/2**2.** n = 3, l = 2, m_{1}= 1, m_{s} =+1/2**3.** n = 4, l= 1, m_{1} = 0, m_{s} = +1/2**4.** n = 3, l = 2,m_{1} = -2, m_{s} = -M2**5.** n = 3, l = 1, m_{1}=-1, m_{s} =+1/2**6.** n = 4, l = 1, m_{1} = 0, m_{s}s =+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, m_{1} = -2, m_{s} = -1/2 (Energy = 4 + 2 = 6)**2.** n = 3, l = 2, m_{1}= 1, m_{s} =+1/2 (Energy = 3 + 2 = 5)**3.** n = 4, l= 1, m_{1} = 0, m_{s} = +1/2 (Energy = 4 + 1 = 5)**4.** n = 3, l = 2,m_{1} = -2, m_{s} = -1/2 (Energy = 3 + 2 = 5)**5.** n = 3, l = 1, m_{1}=-1, m_{s} =+1/2 (Energy = 3 + 1 = 4)**6.** n = 4, l = 1, m_{1} = 0, m_{s}s =+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 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}

Electronic configuration of Si is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{2}

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

Element | Electronic Configuration | Unpaired Electrons |
---|---|---|

P | 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3} | 3 |

Si | 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{2} | 2 |

Cr | [Ar] 3d^{5} 4s^{1} | 6 |

Fe | [Ar] 3d^{6} 4s^{2} | 4 |

Kr | [Ar] 3d^{10} 4s^{2} 4p^{6} | 0 |

**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 2n^{2} = 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

Table of Contents

## 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/n_{1}^{2} – 1/n_{2}^{2})

Where

λ = Wavelength of light emitted by an electron

n_{1} and n_{2} are Integers (n_{2} > n_{1}) both n_{1},n_{2} 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**

E_{n} = -2.18 × 10^{-18}/n^{2} J atom^{-1}

Where

n = Orbit of electron**Formula to calculate radius of any orbit of hydrogen atom is**

r_{n} = 0.0529/n^{2} 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.