In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

1. increasing the filament current

2. decreasing the filament current

3. decreasing the potential difference between the anode and filament

4. increasing the potential difference  between the anode and filament

A radioactive nucleus of mass M emits a photon

of frequency $\nu$ and the nucleus recoils. The recoil

energy will be:

(1) ${\mathrm{h}}^2{\mathrm{\nu }}^2/2 {\mathrm{Mc}}^2$

(2) zero

(3) h$\nu$

(4) ${\mathrm{Mc}}^2-\mathrm{h\nu }$

In photoelectric emission process from a metal

of work function 1.8 eV, the kinetic energy of most

energetic electrons is 0.5 eV. The corresponding

stopping potential is

(1) 1.2 V

(2) 0.5 V

(3)2.3 V

(4) 1.8 V

Light of two different frequencies whose

photons have energies 1 eV and 2.5 eV

respectively illuminate a metallic surface 

whose function is 0.5 eV successively.

Ratio of maximum speeds of emitted 

electrons will be

(1) 1:2

(2) 1:1

(3) 1:5

(4) 1:4

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

The energy of a hydrogen atom in the ground state is \(-13.6~\mathrm{eV}\). What is the energy of a He⁺ ion in the first excited state?

1. -13.6 eV

2. -27.2 eV

3. -54.4 eV

4. -6.8 eV

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: