Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 2.1:

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

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

Answer:

(i) Mass of one electron = 9.10939 × 10^–31 kg

Number of electrons that weigh 9.10939 × 10^–31 kg = 1

Number of electrons that will weigh 1 g = (1 × 10^–3 kg)

= 0.1098 × 10^{–3 + 31}

= 0.1098 × 10²⁸

= 1.098 × 10²⁷

(ii) Mass of one electron = 9.10939 × 10^–31 kg

Mass of one mole of electron = (6.022 × 10²³) × (9.10939 ×10^–31 kg)

= 5.48 × 10^–7 kg

Charge on one electron = 1.6022 × 10^–19 coulomb

Charge on one mole of electron = (1.6022 × 10^–19 C) (6.022 × 10²³)

= 9.65 × 10⁴ C

Question 2.2:

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

(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?

Answer:

(i) Number of electrons present in 1 molecule of methane (CH₄)

{1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 × 10²³ molecules of methane

= 6.022 × 10²³ × 10 = 6.022 × 10²⁴

(ii) (a) Number of atoms of ¹⁴C in 1 mole= 6.023 × 10²³

Since 1 atom of ¹⁴C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of ¹⁴C is (6.023 × 10²³) ×8. Or, 14 g of ¹⁴C contains (6.022 × 10²³ × 8) neutrons.

Number of neutrons in 7 mg

= 2.4092 × 10²¹

(b) Mass of one neutron = 1.67493 × 10^–27 kg

Mass of total neutrons in 7 g of ¹⁴C

= (2.4092 × 10²¹) (1.67493 × 10^–27 kg)

= 4.0352 × 10^–6 kg

(iii) (a) 1 mole of NH₃ = {1(14) + 3(1)} g of NH₃

= 17 g of NH₃

= 6.022× 10²³ molecules of NH₃

Total number of protons present in 1 molecule of NH₃

= {1(7) + 3(1)}

= 10

Number of protons in 6.023 × 10²³ molecules of NH₃

= (6.023 × 10²³) (10)

= 6.023 × 10²⁴

⇒ 17 g of NH₃ contains (6.023 × 10²⁴) protons.

Number of protons in 34 mg of NH³

= 1.2046 × 10²²

(b) Mass of one proton = 1.67493 × 10^–27 kg

Total mass of protons in 34 mg of NH₃

= (1.67493 × 10^–27 kg) (1.2046 × 10²²)

= 2.0176 × 10^–5 kg

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Question 2.3:

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

,  ,  ,  ,

Answer:

¹³₆C:

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) – (Atomic number)

= 13 – 6 = 7

:

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass) – (Atomic number)

= 16 – 8 = 8

:

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass) – (Atomic number)

= 24 – 12 = 12

:

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) – (Atomic number)

= 56 – 26 = 30

:

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass) – (Atomic number)

= 88 – 38 = 50

Question 2.4:

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

Answer:

(i)   (ii)   (iii) 

Question 2.5:

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

Answer:

From the expression,

We get,

 …….. (i)

Where,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of yellow light = 580 nm = 580 × 10^–9 m

Substituting the values in expression (i):

Thus, frequency of yellow light emitted from the sodium lamp

= 5.17 × 10¹⁴ s^–1

Wave number of yellow light, 

Question 2.6:

Find energy of each of the photons which

(i) correspond to light of frequency 3× 10¹⁵ Hz.

(ii) have wavelength of 0.50 Å.

Answer:

(i) Energy (E) of a photon is given by the expression,

E = 

Where,

h = Planck’s constant = 6.626 × 10^–34 Js

ν = frequency of light = 3 × 10¹⁵ Hz

Substituting the values in the given expression of E:

E = (6.626 × 10^–34) (3 × 10¹⁵)

E = 1.988 × 10^–18 J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

Substituting the values in the given expression of E:

Question 2.7:

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^–10 s.

Answer:

Frequency (ν) of light 

Wavelength (λ) of light 

Where,

c = velocity of light in vacuum = 3×10⁸ m/s

Substituting the value in the given expression of λ:

Wave number   of light 

Question 2.8:

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

Answer:

Energy (E) of a photon = hν

Energy (En) of ‘n’ photons = nhν

Where,

λ = wavelength of light = 4000 pm = 4000 ×10^–12 m

c = velocity of light in vacuum = 3 × 10⁸ m/s

h = Planck’s constant = 6.626 × 10^–34 Js

Substituting the values in the given expression of n:

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10¹⁶.

Question 2.9:

A photon of wavelength 4 × 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 × 10^–19 J).

Answer:

(i) Energy (E) of a photon = hν

Where,

h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of photon = 4 × 10^–7 m

Substituting the values in the given expression of E:

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission E_k is given by

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (ν) can be calculated by the expression,

Where,   is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

v = 5.84 × 10⁵ ms^–1

Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^–1.

Question 2.10:

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

Answer:

Energy of sodium (E)

= 4.947 × 10⁵ J mol^–1

= 494.7 × 10³ J mol^–1

= 494 kJ mol^–1

Question 2.11:

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

Answer:

Power of bulb, P = 25 Watt = 25 Js^–1

Energy of one photon, E = hν 

Substituting the values in the given expression of E:

E = 34.87 × 10^–20 J

Rate of emission of quanta per second

Question 2.12:

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

Answer:

Threshold wavelength of radian  = 6800 × 10^–10 m

Threshold frequency  of the metal

= 4.41 × 10¹⁴ s^–1

Thus, the threshold frequency   of the metal is 4.41 × 10¹⁴ s^–1.

Hence, work function (W0) of the metal = hν₀

= (6.626 × 10^–34 Js) (4.41 × 10¹⁴ s^–1)

= 2.922 × 10^–19 J

Question 2.13:

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?

Answer:

The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

Substituting the values in the given expression of E:

E = – (4.0875 × 10^–19 J)

The negative sign indicates the energy of emission.

Wavelength of light emitted 

Substituting the values in the given expression of λ: