Dual Nature of Radiation and Matter · Question 12

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is m. (h = 6.63 10-34 J s, e = 1.6 10-19 C, c = 3 108 m/s)

1. 2.9 10-8
2. 2.5 10-7
3. 3.1 10-7
4. 2.2 10-8

The de Broglie wavelength of an oxygen molecule at 27 is x 10-12 m. The value of x is (take Planck's constant = 6.63 10-34 J s, Boltzmann constant = 1.38 10-23 J/K, mass of oxygen molecule = 5.31 10-26 kg)

1. 24
2. 30
3. 20
4. 26

Light is incident on a metallic plate having work function 110 10-20 J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is rad/s. (h = 6.63 10-34 J s)

1. 1.04 1013
2. 1.66 1016
3. 1.66 1015
4. 1.04 1016

A light wave described by E = 60 [ (3 1015 t ) + (12 1015 t ) ] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) eV. (h = 6.6 10-34 J s and e = 1.6 10-19 C)

1. 3.8
2. 7.8
3. 6.0
4. 5.1

A particle having electric charge 3 10-19 C and mass 6 10-27 kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is 10-12 m. The value of is . (Take Planck's constant = 6.6 10-34 J s)

An electron is released from rest near an infinite non-conducting sheet of uniform charge density '- '. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical value of n is .

A photoemissive substance is illuminated with a radiation of wavelength i so that it releases electrons with de-Broglie wavelength e. The longest wavelength of radiation that can emit photoelectron is o. Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light)

1. e = h i 2mc
2. e = h 2mc ( 1 i - 1 o )
3. e = h2mc ( 1 i - 1 o )
4. e = h o2mc

A small mirror of mass m is suspended by a massless thread of length l. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c = speed of light in vacuum and g = acceleration due to gravity)

1. = Emc gl
2. = E2mc gl
3. = 3E4mc gl
4. = 2Emc gl

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R In the light of the above statements, choose the correct answer from the options given below

1. Both A and R are true and R is the correct explanation of A
2. A is false but R is true
3. Both A and R are true but R is NOT the correct explanation of A
4. A is true but R is false

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

1. Red
2. Green
3. Blue
4. Yellow