If the work function of two metals X and Y are 4 . 17 eV and 8 . 24 × 10 - 19 J respectively, then for which metal lesser amount of energy is required to emit electron?

Given the work function of the metal X = 4 . 17 eV . The work function of metal Y = 8 . 24 × 10 - 19 J = 8 . 24 × 10 - 19 1 . 6 × 10 - 19 = 5 . 15 eV . The minimum amount of energy required by an electron to just escape from the metal surface is called the work function of the metal. It is clear that the metal X has a lower work function hence it requires less amount of energy to eject electrons from its surface.

The threshold frequency for a certain metal is 3 . 3 × 10 14 Hz . If the light of frequency 8 . 2 × 10 14 H z is incident on the metal, predict the cut-off voltage for the photoelectric emission. ( h = 6 . 62 × 10 - 34 m 2 kg / s )

Given the threshold frequency of the metal= 3 . 3 × 10 14 Hz . The frequency of incident light = 8 . 2 × 10 14 Hz . Using Einstein's photoelectric equation, Cut-off voltage for the photoelectric emission from the metal = V 0 is given by, 1 2 mv max 2 = h ( ν = ν 0 ) = eV 0 or V 0 = h ν - ν 0 e . Substituting the values, we get, V 0 = 6 . 62 × 10 - 34 1 . 6 × 10 - 19 8 . 2 × 10 14 - 3 . 3 × 10 - 14 = 2 . 03 eV . Hence, the threshold voltage is 2 . 03 eV

The threshold frequency for a certain metal is 3 . 3 × 10 14 Hz . If the light of frequency 8 . 2 × 10 14 H z is incident on the metal, predict the cut-off voltage for the photoelectric emission. ( h = 6 . 62 × 10 - 34 m 2 kg / s )

Given the threshold frequency of the metal= 3 . 3 × 10 14 Hz . The frequency of incident light = 8 . 2 × 10 14 Hz . Using Einstein's photoelectric equation, Cut-off voltage for the photoelectric emission from the metal = V 0 is given by, 1 2 mv max 2 = h ( ν = ν 0 ) = eV 0 or V 0 = h ν - ν 0 e . Substituting the values, we get, V 0 = 6 . 62 × 10 - 34 1 . 6 × 10 - 19 8 . 2 × 10 14 - 3 . 3 × 10 - 14 = 2 . 03 eV . Hence, the threshold voltage is 2 . 03 eV

If the work function of two metals X and Y are 4 . 17 eV and 8 . 24 × 10 - 19 J respectively, then for which metal lesser amount of energy is required to emit electron?

Given the work function of the metal X = 4 . 17 eV . The work function of metal Y = 8 . 24 × 10 - 19 J = 8 . 24 × 10 - 19 1 . 6 × 10 - 19 = 5 . 15 eV . The minimum amount of energy required by an electron to just escape from the metal surface is called the work function of the metal. It is clear that the metal X has a lower work function hence it requires less amount of energy to eject electrons from its surface.

What is de Broglie wavelength associated with(a) an electron of mass 9 . 11 × 10 - 31 kg moving with velocity 5 . 4 × 10 6 m / s and a ball of mass 150 g travelling at 30 m / s . Given h = 6 . 63 × 10 - 34 J - s

Given the mass of the electron = m e = 9 . 11 × 10 - 31 kg . The velocity of the electron v e = 5 . 4 × 10 6 m / s . The mass of the ball m b = 150 g . = 0 . 15 kg . The speed of the ball v b = 30 m / s . Planck's constant h = 6 . 6 × 10 - 34 Js The de Broglie wavelength of a particle is given by λ = h mv . Therefore, for electrons, we have, λ e = 6 . 6 × 10 - 34 9 . 11 × 10 - 31 × 5 . 4 × 110 6 = 0 . 135 nm ......(1) For the ball, we get, λb = 6 . 6 × 10 - 34 0 . 15 × 30 = 1 . 47 × 10 - 34 m .....................(2) Therefore, the de Broglie associated with the electron = 0 . 135 nm and with the ball is 1 . 47 × 10 - 34 m .

A laser light emits light of the frequency of 6 × 10 14 Hz and power emitted is 2 × 10 - 3 W . How many photons per second on an average are emitted by the source. ( h = 6 . 6 × 10 - 34 JS )

Given the power of the source P = 2 × 10 - 3 W . The frequency of the radiation ν = 6 × 10 14 Hz . The energy of the radiation is given by Planck's relation: E = hν = 6 . 6 × 10 - 34 × 6 × 10 14 , where h = Planck's constant. The number of photons emitted is given by n = power energy = 2 × 10 - 3 6 . 6 × 10 - 34 × 6 × 10 14 = 5 × 10 15 per sec . Hence, the number of photons emitted per second= 5 × 10 15 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2018>Q 3

EASY

12th Assam Board

IMPORTANT

Earn 100

What determines the intensity of light in the photon picture of light?

Important Questions on Dual Nature of Radiation and Matter

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2019>Q 4

If the work function of two metals

X and Y

are

4.17 eV and 8.24 \times 10^{- 19} J

respectively, then for which metal lesser amount of energy is required to emit electron?

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2020>Q 5

Guess the shape of curve which shows the variation of $v_{\circ}$ with v in case of photoelectric emission shown by the relation where symbols have their usual meaning.

$V_{\circ} = (\frac{h}{e}) v - \frac{ϕ_{\circ}}{e}$

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2020>Q 6

The de Borgile wavelength of a heavier particle is _____.

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2020>Q 7

What did de Broglie propose regarding wavelength $(λ)$ associated with a particle of mass $m$ ?An electron of mass $m$ and charge $e$ is accelerated from rest through potential $V$ . Show that the deBroglie wavelength of the electron is,

$V = \frac{1.227}{\sqrt{V}}$

Given $\frac{h}{\sqrt{2 m e}} = 1.227$

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 11 - Dual Nature of Radiation and Matter>assam board-2020>Q 8

The threshold frequency for a certain metal is $3.3 \times 10^{14} Hz$ . If the light of frequency $8.2 \times 10^{14} H z$ is incident on the metal, predict the cut-off voltage for the photoelectric emission. ( $h = 6.62 \times 10^{- 34} m^{2} kg / s$ )

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