Which of the following figures represents the variation of the particle momentum and the associated de-Broglie wavelength?

1.		2.
3.		4.

A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material will be:
(h = Planck’s constant, c = speed of light)

1. hc/2λ

2. hc/λ

3. 2hc/λ

4. hc/3λ

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5 eV to 0.8 eV. The work function of the metal will be:
1. 0.65 eV

2. 1.0 eV

3. 1.3 eV

4. 1.5 eV

What will be the percentage change in the de-Broglie wavelength of the particle if the kinetic energy of the particle is increased to 16 times its previous value?
1. 25
2. 75
3. 60
4. 50

For photoelectric emission from certain metal, the cut-off frequency is $\nu$. If radiation of frequency 2$\nu$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

1. $\sqrt{h\nu /\left(2m\right) }$

2. $\sqrt{h\nu /\left(m\right) }$

3. $\sqrt{2h\nu /\left(m\right) }$

4. none of these

Photoelectric emission occurs only when the incident light has more than a certain minimum 

1. wavelength

2. intensity

3. frequency

4. power

Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would

1. decrease by 2 times

2. decrease by 4 times

3. increase by 4 times

4. increase by 2 times

The threshold frequency for a photo-sensitive metal is $3.3\times {10}^{14} \mathrm{Hz}.$ If the light of frequency $8.2\times {10}^{14} \mathrm{Hz}$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly:

1. 2 V                                        2. 3 V

3. 5 V                                         4. 1 V

A source S₁ is producing, 10¹⁵ photons/s of wavelength 5000 $\stackrel{\circ }{A}$. Another source S₂ is producing 1.02$\times$ 10¹⁵ photons per second of wavelength 5100 $\stackrel{\circ }{A}$. Then, (power of S₂)/(power of S₁) is equal to

1. 1.00                 2. 1.02

3. 1.04                 4. 0.98

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface having a work function of 5.01 eV when ultraviolet light of 200 nm falls on it is: