Dual Nature of Radiation and Matter: Important Questions

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Dual nature of radiation and matter is basically the study of different natures that a matter possesses or exhibits. Photons are the packets of energy that serve as the quantum of the electromagnetic field, including electromagnetic radiation that is emitted by a source of radiation. NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter are provided in the article

Photons travel in a straight line with the speed of light.
As per Plank’s Quantum Theory, a source of radiation emits energy in the form of photons that travel in a straight line.
The frequency of a photon gives the radiation definite energy (or color), which does not change when photons travel through different media.
In a different medium, there is a change in the velocity of a photon which is due to a change in its wavelength.

Very Short Answer Questions [1 Mark Questions]

Ques. What is the value of sparking potential for air at normal pressure?

Ans. The value of sparking potential at normal pressure is 30,000 V.

Ques. Who discovered the electron?

Ans. Electron was discovered by J.J. Thomson in 1887.

Ques. Which all properties do cathode rays dependent upon?

Ans. Following are the properties on which cathode ray depends

Nature of gas in the discharge tube
Nature of material of the cathode

Ques. Who discovered the photoelectric effect?

Ans. The photoelectric effect was discovered by Heinrich Hertz.

Ques. What is the rest mass of a photon? (CBSE 1990)

Ans. Zero

Ques. If h is Plank’s constant, find the momentum of a photon of wavelength 0.02 A°. (CBSE 1990)

Ans. Momentum of photon = h/λ = h / 0.02 x 10 ^-10 = 50 x 10 ¹⁰ h

Ques. Do non-metals show a photo-electric effect?

Ans. Yes, they show a photo-electric effect with light of high frequency.

Short Answer Questions [2 Marks Questions]

Ques. Is photoelectric emission possible at all frequencies? Give reasons for your answer. (CBSE 1990)

Ans. The photoelectric emission is possible if the energy of the incident photon is more than the work function of the metal i.e., the energy required to just liberate an electron from the surface of the metal. If the energy place of the incident photon is less than the work functions of a metal from a metal surface.

Ques. Explain the effect of the increase of i) frequency and ii) intensity of the incident radiation on photo-electrons emitted by a photo tube. (CBSE 1994)

Ans. i) With the increase of frequency of incident radiation, K.E. of the emitted photoelectric increases.

ii) With the increase of intensity of the incident radiation, the number pf photoelectric emitted per unit time increases.

Ques. State two laws of photoelectric emission. Is cathode rays wave or particle?

Ans. Cathode rays are not waves but streams of fast-moving electrons.

Ques. What is the momentum of a photon of wavelength 0.01°A

Ans. p = h/λ = h /0.01x 10^-10 = 10¹²h

Ques. Why a photo-electric cell is also called an electric eye?

Ans. The photo-electric cell is also called as electric eye as the photoelectric current setup in the photoelectric cell corresponds to incident lights that provide information about the objects as seen by our eyes in the presence of light.

Long Answer Questions [3 Marks Questions]

Ques. Describe the photoelectric effect and state the laws of photoelectric emission. (HBSE 2000)

Ans. The photoelectric Effect is the phenomenon of emission of electrons from the surface of metals when radiations of suitable frequency fall on them. The emitted electrons are photo-electrons and the current so produced is called photoelectric current.

The photoelectric effect was discovered by Hallwachs experimentally verified by Hertz and successfully explained theoretically by Einstein.

Ques. Describe the laws of the photoelectric effect.

Ans. The laws of photoelectric effects are as below:

The given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light.
With a given metal, there exists some certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is called threshold frequency.
With a threshold frequency set, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light but depends only upon the frequency (or wavelength) of the incident light.
Photoelectric emission is an instantaneous process. The time lag between the incidence of radiations and emissions of photoelectrons is very small, less than even 10^-9seconds.

Ques. Calculate the frequency associated with a photon of energy 3.3 x 10^-20 J (h = 6.6 x 10-³⁴Js) (CBSE 1990)

Ans. E = 3.3 x 10^-20 J, h = 6.6 x 10^-34 Js

Applying the Formula De Broglie Dualistic Hypothesis

E = hη

η = E/h

η = 3.3 x 10 ^-20 / 6.6 x 10 ^-34

η = 5 * 10 ¹³ Hz

Very Long Answer Questions [5 Marks Questions]

Ques. The threshold frequency for a certain 3.3 x 10¹⁴ Hertz. If the light frequency is 8.2 x 10¹⁴Hz. If the light of frequency 8.2 x 10 ¹⁴ Hz is incident on the metal, predict the cut-off voltage for photoelectric emission. Given h = 6.63 x 10 ^{– 34} Js ; e= 1,6 x 10 ^{– 19} C. ( NCERT)

Ans. Here v₀ = 3.3 x 10¹⁴ Hertz , v = 8.2 x 10 ¹⁴ Hz

By the relation between cut-off potential, frequency of the incident photon, and Threshold frequency

E v₀ = hv – hv₀

_v0 = h ( v – v₀) / e

= 6.63 x 10 ^{– 34} ( 8.82 x 10 ^-34 - 3.3 x 10¹⁴ ) / 1.6 x 10^-19)

= 2.03 V

Ques. The minimum length intensity that can be received by the eye is about 10^-10 Wm^-2. Find the number of photos of wavelength 5.84 x 10^-7 m that must enter the pupil, of area 10^-4 m² s^-1 for vision, Given h – 6.6 x 10 ^-34 Js?

Ans. Applying the formula De Broglie Dualistic Hypothesis

Applying the Formula De Broglie Dualistic Hypothesis

E = hη

E = hc / λ

⇒ n hc / λ = IA

n = λ IA / hc

= 10 ^-10 x 10 ^-4 x 5.84 x 10 ^-7 / 6.6 x 10 ^-34 x 3 x 10 ⁸

= 3 x 10 ⁴

Ques. a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength ending at 0.45° A. What is the maximum energy of a photon in the radiation?

b) From your answer to (a), give what order of accelerating voltage (for electrons) is required in such a tube. (NCERT)

Ans. (a) λ_min = 0.45 °A

= 0.45 x 10^-10 m

h = 6.63 x 10^-34 Js

c = 3 x 10⁸ms^-1

We know, the maximum energy of an X-ray photon is

E _max = h λ_max = hc/ λ_min

= 6.63 x 10^-34 x 3 x 10⁸ / 0.45 x 10^-10 J

= 27.6 x 10³ eV

= 27.6 keV

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CBSE CLASS XII Related Questions

1.
(a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
% Given data Given:
\( m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} \)
\( m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} \)
\( m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} \)
\( m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} \)
\( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)
View Solution
2.
Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), \( r \) being the separation between the two charges \( (q_1, q_2) \) and \( (q_2, q_3) \). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1, q_2 \), and \( q_3 \) are such that:
- \( q_2<q_1<q_3 \)
- \( q_3<q_1<q_2 \)
- \( q_1<q_2<q_3 \)
- \( q_2<q_3<q_1 \)
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3.
Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the \( n \)-th orbit,
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).
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4.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
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5.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:
- zero
- less than \( g \)
- \( g \)
- greater than \( g \)
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6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).
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