# NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 11 Dual Nature of Radiation and Matter (Part II)

NCERT Solutions for 12th Physics Chapter 11, Dual Nature of Radiation and Matter are available here. This chapter is a continuation of Part I (same chapter) where solutions from question number 11.1 to 11.10 are available. In this part, you will get solutions from question number 11.11 to 11.19. All the questions of this chapter is extremely important for CBSE 12th board exam 2018 and other competitive exams.

NCERT Solutions for Class 12 Physics ‒ Chapter 11: Dual Nature of Radiation and Matter from question number 11.11 to 11.19 are given below:

Question 11.11: Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

Solution 11.11:

Given: 𝜆= 488 nm

Vo = 0.38 V

We know

eVo = hv – Φo

Or

Φo = hc / 𝜆– eVo

= 3.46 x 10‒19 J

= 2.16 eV.

NCERT Exemplar Questions: CBSE Class 12 Physics – Chapter 11

Question 11.12: Calculate the

(a) momentum, and

(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Solution 11.12:

(a) Using the relation,

P = (2 m e V)1/2 = 4.04 × 10‒24 kg m/s.

(b) 𝜆= h / p = 0.164 nm.

Question 11.13: What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV. Given, h = 6.63 × 10‒34 Js, me = 9 × 10‒31 kg, 1 eV = 1.6 × 10‒19 J.

Solution 11.13:

(a) p2 = 2 m K

Or p = 5.91 × 10‒24 kg m/s

(b) Speed can be calculated by using,

mv2 / 2 = K

mv2 / 2 = 1.92 × 10-17

v = 6.5 × 106 m / s

(c) 𝜆= h/p = 0.112 nm.

Question11.14: The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which

(a) an electron, and

(b) a neutron, would have the same de Broglie wavelength.

Solution 11.14:

Given: 𝜆= 589 nm

For this 𝜆,

p = h / 𝜆 = 1.12 × 10-17 kg m / s

(a) For electron,

K = p2 / 2me = 6.9 × 10-25 J

(b) For neutron,

K = p2 / 2mn = 3.7 × 10-28 J.

Question 11.15: What is the de Broglie wavelength of

(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and

(c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s?

Solution 11.15:

(a) m = 0.04 kg

v = 1 km/s

𝜆= h / mv = 1.66 x 10-35 m

(b) m = 0.06 kg

v = 1 m/s, 𝜆= 1.1 x 10-32 m.

(c) m = 1 x 10-9 kg

v = 2.2 m/s

𝜆=3 x 10-25 m

Question 11.16: An electron and a photon each have a wavelength of 1.00 nm. Find

(a) Their momenta,

(b) the energy of the photon, and

(c) the kinetic energy of electron.

Solution 11.16:

(a) Momentum of electron

p = h / λ = 6.63 x 10-25 kg m / s

Momentum of photon

p = h / λ = 6.63 x 10-25 kg m / s

(b) Energy of photon

E = mc2 …(1)

Now,

λ = h / mc

Or m = h / cλ

Substituting value of m in eq. (1)

E = hc / λ = 1.24 keV

(c) Kinetic energy of electron, K = p2 / 2m = 2.41 x 10-19 J = 1.51 eV.

Question 11.17: (a) For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.40 × 10–10m?

(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

Solution 11.17:

Given: 𝜆= 1.4 x 10‒10 m

(a) Now

K = p2 / 2m and p = h / 𝜆. Therefore, K = h2 / 2m𝜆2 or K = 6.686 × 10‒21 J

(b) Kinetic Energy

K = (3 / 2) kT

Where,

k = Boltzmann constant = 1.381 × 10‒23 J / K

T = 300 K

So,

K = 6.21 x 10‒21 J.

Now, λ = h /(2 m k)1/2 = 0.145 nm.

Question 11.18: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Solution 11.18:

For a photon,

p = hν /c

Therefore,

h / p = c / v = λ

That is, the de Broglie wavelength of a photon equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum.

Question 11.19: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Solution 11.19:

Sol.

Given:

T = 300 K

m = 14.0076 x 1.661 x 10‒27 kg

v = rms speed of molecules at T

K = 1.5 kT

Substitution results,

K = 6.21 x 10‒21 J

Now,

λ = h / (2mK)1/2

On putting values and solving, we have, λ = 0.038 m.

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