In the Davisson and Germer experiment, how can we increase the velocity of electrons emitted from the electron gun?

The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three difference radiations. Which one of the following is a correct statement?

(1) Curves a and b represent incident radiations of different frequencies and different intensities

(2) Curves a and b represent incident radiations of same frequency but of different intensities 

(3) Curves b and c represent incident radiations of different frequencies and different intensities

(4) Curves b and c represent incident radiations of same frequency having same intensity 

The number of photoelectrons emitted for light of a frequency v (higher than the threshold frequency $v_0$) is proportional to

1. $v-v_0$

2. threshold frequency $\left(v_0\right)$

3. intensity of light

4. frequency of light (v)

The de-Broglie wavelength of a particle moving with a velocity $2.25\times {10}^8$ m/s is equal to the wavelength of the photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3\times {10}^8$ m/s)

(1) 1/8                   

(2) 3/8
(3) 5/8                   

(4) 7/8

The kinetic energy of electron and proton is ${10}^{-32}$ J. Then the relation between their de-Broglie wavelengths is

(1) ${\mathrm{\lambda }}_{\mathrm{p}}<{\mathrm{\lambda }}_{\mathrm{e}}$                 

(2) ${\mathrm{\lambda }}_{\mathrm{p}}>{\mathrm{\lambda }}_{\mathrm{e}}$
(3) ${\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{e}}$                 

(4) ${\mathrm{\lambda }}_{\mathrm{p}}=2{\mathrm{\lambda }}_{\mathrm{e}}$

The de-Broglie wavelength of a particle accelerated with 150 volt potential is ${10}^{-10}$ m. If it is accelerated by 600 volts p.d., its wavelength will be
(1) 0.25 Å                 

(2) 0.5 Å
(3) 1.5 Å                   

(4) 2 Å

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of 3 km/s will be
(1) 1 Å                     

(2) 0.66 Å

(3) 6.6 Å                 

(4) 66 Å

When the momentum of a proton is changed by an amount ${\mathrm{P}}_0$, the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was 
(a) ${\mathrm{P}}_0$                   (b) 100  ${\mathrm{P}}_0$
(c) 400 ${\mathrm{P}}_0$             (d) 4 ${\mathrm{P}}_0$

The de-Broglie wavelength of a neutron at 27 $°\mathrm{C}$ is $\mathrm{\lambda }$. What will be its wavelength at 927 $°\mathrm{C}$
(a) $\mathrm{\lambda }$ / 2                        (b) $\mathrm{\lambda }$ / 3
(c) $\mathrm{\lambda }$ / 4                        (d) $\mathrm{\lambda }$ / 9

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

(1) Zero
(2) Infinity
(3) Equal to the kinetic energy of the proton
(4) Greater than the kinetic energy of the proton