# Atomic Structure

## NCERT Solution

### Part 1

Question 1:

(a) Calculate the number of electrons which will together weight one gram.

Answer: Mass of 1 electron = 9.1094 × 10-31 kg

= 9.1094 × 10-28 g

So, number of electrons in 1 g = (1 g) ÷ (9.1094 × 10-28 g)

= 1.1 × 1027

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

Answer: Mass of 1 electron = 9.1094 × 10-31 kg

So, mass of 1 mole electrons = 6.022 × 1023 × 9.1094 × 10-31 kg

= 5.486 × 10-7 kg

Charge on 1 electron = -1.602176 × 10-19 C

So, charge on 1 mole electrons = 6.022 × 1023 × -1.602176 × 10-19

= 9.64 × 104

Question 2:

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

Answer: Number of electrons present in one molecule of methane (CH4) = 6 + 4 = 10

So, number of electrons present in 1 mole of methane = 6.022 × 1023 × 10

= 6.022 × 1024

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

Answer: We know that atomic number of C = 6

So, no. of protons in C = 6

So, no. of neutrons in C = 14 – 6 = 8

As mass number = 14

So, 14 g of given C contains 1 mole carbon

No. of neutrons in 1 mole (or 14 g) of given C = 8 × 6.022 × 1023

So, number of neutrons in 7 g C = (8 × 6.022 × 1023) ÷ 2

So, number of neutrons 7 mg C = (2 × 6.022 × 1023) ÷ 1000

= 1.204 × 1021

Now, total mass of neutron = 1.675 × 10-27 kg × 1.204 × 1021

= 2.016 × 10-6 kg

(c) Find the total number and the total mass of protons in 34 mg of NH3 at STP.

Will the answer change if the temperature and pressure are changed?

Answer: Number of protons in NH3 = 7 + 3 = 10

Molecular mass of NH3 = 14 + 3 = 17

So, number of molecules in 17 g ammonia = 6.022 × 1023

Number of protons in 17 g (or 17000 mg) ammonia = 10 × 6.022 × 1023

So, no. of protons in 34 mg ammonia = 10 × 6.022 × 1023 × 34 ÷ 17000

= 1.202 × 1022

Now, mass of 1 proton = 1.67267 × 10-27 kg

So, mass of given number of protons

= 1.67267 × 10-27 kg × 1.202 × 1022

= 2.01 × 10-5 kg

Question 3: How many neutrons and protons are there in the following nuclei?

136C, 168O, 2412Mg, 5626Fe, 8838Sr

Answer: Atomic number (Z) of C = 6

Mass number (A) of C = 13

So, no. of protons in C = 6

No. of neutrons in C = 13 – 6 = 7

Atomic number (Z) of O = 8

Mass number (A) of O = 16

So, no. of protons in O = 8

No. of neutrons in O = 16 – 8 = 8

Similarly,

No. of protons in Mg = 12 and no. of neutrons = 12

No. of protons in Fe = 26 and no. of neutrons = 30

No. of protons in Sr = 38 and no. of neutrons = 50

Question 4: Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

1. Z = 17, A = 35
2. Z = 92, A = 233
3. Z = 4, A = 9

Answer: (a) Cl, (b) U, (c) Be

Question 5: Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν) of the yellow light.

Answer: Frequency of yellow light can be calculated as follows:

We know that ν = c/λ

Given: c = 3 × 108 ms-1

λ = 580 nm = 580 × 10-9 m

So, ν =(3xx10^8m\s^(-1))/(580×10^(-9)m)

= 5.17 × 10-14 s-1

Wave number of yellow light can be calculated as follows:

Wave number (ν) =1/λ=1/(580xx10^(-9)m)

= 1.724 × 106 m-1

Question 6: Find the energy of each of the photons which

(a) Correspond to light of frequency 3 × 1015 Hz.

Answer: Energy of photon (E) = hν

We know that Planck’s constant h = 6.626 × 10-34 J s

Given ν = 3 × 1015 Hz = 3 × 1015 s-1

So, E = (6.626 × 10-34 J s) × (3 × 1015 s-1)

= 1.9878 × 10-18 J

(b) Have wavelength of 0.50 Å.

Answer: Frequency can be calculated as follows:

ν = c/λ

= (3xx10^8m\s^(-1))/(0.5xx10^(-10)m)=6xx10^(18) s-1

Now, energy can be calculated as follows:E = hν

= (6.626 × 10-34 J s) × (6 × 1018 s-1)

= 3.975 × 10-15 J

Question 7: Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10-10 s.

= 1 ÷ (2.0 × 10-10 s)

= 0.5 × 1010 s-1

Now, wavelength can be calculated as follows:

λ=c/ν

= (3 × 108 ms-1) ÷ (0.5 × 1010 s-1)

= 6 × 10-2 m

Wavenumber (ν) = 1/λ

= 1 ÷ (6 × 10-2 m)

= 16.67 m-1

Question 8: What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Answer: Wavelength λ = 4000 pm = 4000 × 10-12 m

= 4 × 10-9 m

Planck’s constant: h = 6.626 × 10-34 J s

Speed of light c = 3 × 108 m s-1

Energy of photon(E) = (hc)/(λ)

=(6.626xx10^(-34)Js\xx3xx10^8ms^(-1))/(4xx10^(-9)m)

= 4.97 × 10-17 J

Question 9: A photon of wavelength 4 × 10-7 m strikes a 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 × 10-19 J).

Answer: Energy of photon (E) =(hc)/(λ)

=(6.626xx10^(-34)Js\xx3xx10^8ms^(-1))/(4xx10^(-7)m)

= 4.97 × 10-19 J

It can be converted into eV as follows:

=(4.97xx10^(-19)J)/(1.602xx10^(-19)J)

= 3.1 eV

Kinetic energy of emission = E – work function (i.e. KE of emitted electrons)

= 3.1 – 2.13 = 0.97 eV

Velocity of photoelectron can be calculated as follows:

0.97 eV = 0.97 × 1.602 × 10-19 kg m2 s-2

We know that KE = ½ mv2

So, v2 = (2xx0.97xx1.602xx10^(-19)kg\m^2s^(-2))/(9.1xx10^(-31)kg)

= 0.34 × 1012 m2 s-2

Or, v = 5.83 × 105 m s-1

Question 10: Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJ mol-1.

Answer: Wavelength = 242 nm = 242 × 10-9 m

Speed of light c = 3 × 108 m

Planck’s constant h = 6.626 × 10-34 J s

Energy E = (hc)/(λ)

=(6.626xx10^(-34)Js\xx3xx10^8ms^(-1))/(242xx10^9m)

= 0.821 × 10-17 J

Ionization energy per mol (E)

=(0.0821xx10^(-17)J\xx6.022xx10^(23)text(mol)^(-1)J)/(1000)

= 494 kJ mol-1

Question 11: A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 µm. Calculate the rate of emission of quanta per second.

Answer: Wavelength = 0.57 × 10-6 m

Energy of one photon (E) =(hc)/(λ)

=(6.626xx10^(-34)Js\xx3xx10^8ms^(-1))/(0.57xx10^(-6)m)

= 3.48 × 10-19 J

Rate of emission of quanta per second = Power/Energy

Power P = 25 watt = 25 J s-1

So, rate of emission

=(25 Js^(-1))/(3.48xx10^(-19)J)

= 7.18 × 1019 s-1

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 (ν0) and work function (w0 of the metal.

Answer: Threshold frequency ν0 = c/λ

=(3xx10^8ms^(-1))/(68xx10^(-8)m)

= 4.41 × 1014 s-1

Work function W0 = hν0

= (6.626 × 10-34 J s) × (4.41 × 1014 s-1)

= 2.92 × 10M-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?

Wave number (ν) = R_H[1/n_1^2-1/n_2^2]cm^(-1)

Here, n1 = 2, n2 = 4

RH (Rydberg constant) = 109677 cm-1

So, ν =109677[1/(2^2)-1/(4^2)]cm^(-1)

=(109677xx3)/(16)cm^(-1)

Now, λ=1/ν=(16)/(109677xx3)cm

=(16xx10^7)/(109677xx3)nm=486 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 orbit).

Answer: Energy of hydrogen atom in stationary state is given by following equation:

E_n=-R_H(1/(n^2))

Where; n = 1, 2, 3, 4, (orbit number)

RH = 2.18 × 10-18 J

So, ionisation energy for hydrogen electron in orbit n = 5 can be calculated as follows:

IE5 = E - E5

=0-((-2.18xx10^(-18))/(25)) J atom-1

= 8.72 × 10-20 J atom-1

Ionisation energy for hydrogen electron in orbit n = 1

IE1 = E - E1

=0-((-2.18xx10^(-18))/(1)) J atom-1

= 2.18 × 10-18 J atom-1

Comparison (IE_1)/(IE_1)=25

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?

Answer: Maximum number of emission lines

=(n(n-1))/2=(6(6-1))/2=15

Question 16:

(a) 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?

Answer: It can be solved as question 14

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

So, r5 = 0.529 × 25 = 13.225 Å

Question 17: Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

ν = 1/λ=R_H[1/(n_1^2)-1/(n_2^2)]

In order to have the longest wavelength, the wavenumber should be the least. It is possible when n2 - n1 is minimum. So, for Balmer series n1 must be 2 and n2 must be 3.

So, ν =(1.097xx10^7 m^(-1))(1/(2^2)-1/(3^2))

=1.097xx10^7m^(-1)(5/(36))

= 1.5233 × 106 m-1