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

NCERT Solutions for 12th Physics Chapter 11, Dual Nature of Radiation and Matter are available here. In this article, you will get solutions from question number 11.1 to 1.10. Solutions of other questions are available in Part II. Some important topics of this chapter are electron emission, photoelectric effect, experimental study of the photoelectric effect, photoelectric effect and wave theory of light, Einstein’s photoelectric equation quantum of radiation, particle nature of light: the photon, wave nature of matter, Davisson and Germer experiment. Most of the questions given in this chapter of NCERT textbook are based on topics mentioned above. This chapter is extremely important for CBSE 12th board exam 2018.

*NCERT Solutions for Class 12 Physics ‒ Chapter 11: Dual Nature of Radiation and Matter from question number 11.1 to 11.10 are given below:*

**Question 11.1:** Find the

**( a)** maximum frequency, and

**( b)** Minimum wavelength of X-rays produced by 30 kV electrons.

**Solution 11.1:**

(*a*) We know,

K_{max} = hv – Φo

Or eV_{o }= hv - Φo

*v*_{max }= eVo / h

Substituting the required values

v_{max }= [(1.6 × 10^{‒19}).(30000)] / (6.63 × 10^{‒34}) = 7.24 × 10^{18} Hz

(*b*) By the relation 𝜆= c / *v*_{max} = 0.041 nm

**NCERT Exemplar Questions: CBSE Class 12 Physics – Chapter 11**

**Question 11.2:** The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10^{14}Hz is incident on the metal surface, photoemission of electrons occurs.

**( a)** What is the maximum kinetic energy of the emitted electrons,

**( b)** Stopping potential, and

**( c)** Maximum speed of the emitted photoelectrons?

**Solution 11.2:**

Given: Φ_{o} = 2.14 V

*v* = 6 × 10^{14 }Hz

(*a*) Maximum kinetic energy is the difference between the photon energy and the work function of the metal. For cesium it is given to be 2.14 V.

Using the expression,

K_{max} = hv ‒ Φ_{o}

Substitution yields,

K_{max} = [(6.63 × 10^{‒34}).(6 × 10^{14})] – [(2.14).(1.6 × 10^{‒19})] = 5.54 × 10^{‒20} J = 0.346 eV.

(*b*) Stopping potential can be deduced by maximum kinetic energy, by the expression

eV_{o }= Kmax

Or V_{o }= *K*_{max }/ *e*

Putting the required values

V_{o} = (5.54 × 10^{‒20}) / (1.6 × 10^{‒19}) = 0.34 V

(*c*) Maximum speed of photoelectron is given by the maximum kinetic energy it possesses,

K_{max} = 5.54 × 10^{‒20}

Or *mv*^{2} / 2 = 5.54 × 10^{‒20}

Or *v*_{max }= 344 km /s.

**Question 11.3:** The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

**Solution 11.3:**

Maximum kinetic energy,

*K*_{max} = eV_{o}

Putting values and solving we have, 2.4 × 10^{‒19} J.

**Question 11.4:** Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.

**( a)** Find the energy and momentum of each photon in the light beam,

**( b)** How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and

**( c)** How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

**Solution 11.4:**

Given:

Wavelength of monochromatic light 𝜆= 632.8 nm

Power of He-Ne laser, P = 9.42 mW

(*a*) Energy of a photon is given by, *E* = *hv*

Or E = *hc* / 𝜆 which gives, E = 3.14 × 10^{‒19 }J

Now momentum of a photon

p = h / 𝜆 Substitution yields

p = 1.05 × 10^{‒27} kg m / s

(*b*) For a beam of uniform cross-section having cross-sectional area less than target area

P = E × N

Where,

P = power emitted

E = energy of photon

N = number of photons

Therefore

N = P / E

Substitution gives

N = 3 × 10^{16} photons / second

(*c*) Momentum of He-Ne laser = 1.05 × 10^{‒27} kg m / s

For this much momentum of a hydrogen atom

*mv *= 1.05 × 10^{‒27}

Or *v* = 1.05 x 10-27 / 1.6 x 10-27

Or *v *= 0.63 m / s

The required speed for hydrogen atom is 0.63 m / s.

**Question 11.5:** The energy flux of sunlight reaching the surface of the earth is1.388 × 10^{3 }W/m^{2}. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

**Solution 11.5:**

Given:

Energy flux = 1388 W / sq. m

Wavelength = 550 nm

Energy of photon = hc / 𝜆 = 3.61 × 10^{‒19} J

So, number of photons = P / E = 4 × 10^{21} photons / m^{2}-s.

**Question 11.6:** In an experiment on photoelectric effect, the slope of the cutoff voltage versus frequency of incident light is found to be 4.12 × 10^{–15} V s. Calculate the value of Planck’s constant.

**Solution 11.6:**

Slope = 4.12 × 10^{-15}

Or *h*/*e* = 4.12 × 10^{-15}

Putting value of *e* and solving,

h = 6.592 ×10^{-34} J s.

**Question 11.7:** A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (*a*) What is the energy per photon associated with the sodium light? (*b*) At what rate are the photons delivered to the sphere?

**Solution 11.7:**

Given:

P = 100 W 𝜆= 589 nm

(*a*) Energy per photon = hc / 𝜆 = 3.37 × 10^{-19} J

(*b*) No. of photons per second = P / E = 3 ×10^{20} photons/second.

**Question 11.8:** The threshold frequency for a certain metal is 3.3 × 10^{14} Hz. If light of frequency 8.2 × 10^{14 }Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

**Solution 11.8:**

Given:

*v*_{o} = 3.3 × 10^{14} Hz

*v* = 8.2 × 10^{14 }Hz

Cut-off voltage

*e*V_{o} = *h *(*v* – *v*_{o})

Substituting the values

V_{o} = 2.03 V.

**Question 11.9:** The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

**Solution11.9:**

𝜙_{o} = 4.2 eV

Or *v*_{o}h = 4.2

Or *v*_{o}= 1.01 × 10^{15} Hz

Incident wavelength = 330 nm or v = 6 × 10^{‒19} Hz

Since *v* is less than *v*_{o}, there will be no photoelectric emission.

**Question** **11.10: **Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10^{5} m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

**Solution11.10:**

Given:

*v* = 7.21 × 10^{14} Hz

v = 6 × 10^{5} m / s

Kinetic energy = h (*v* – *v*o)

Or *mv*^{2} / 2 = h (*v* – *v*_{o})

Solving the above equation

*v*_{o} = 4.73 × 10^{14} Hz.