The number of photoelectrons emitted for the light of a frequency $\nu$ (higher than the threshold frequency $\nu$₀) is proportional to

1. $\nu$ -$\nu$₀
2. threshold frequency ($\nu$₀)
3. intensity of light
4. frequency of light ($\nu$)

A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 x 10⁶ ms^-1. The velocity of the particle is :
(Mass of electron = 9.1 x 10^-31 kg)

1.  $2.7\times {10}^{-18} {\mathrm{ms}}^{-1}$

2.  $9\times {10}^{-2} {\mathrm{ms}}^{-1}$

3.  $3\times {10}^{-31} {\mathrm{ms}}^{-1}$

4.  $2.7\times {10}^{-21} {\mathrm{ms}}^{-1}$

When photons of energy \(h\nu\) fall on an aluminium plate (of work function \(E_0\)), photoelectrons of maximum kinetic energy \(K\) are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:
1. \(K+ E_0\)
2. \(2K\)
3. \(K\)
4. \(K + h\nu\)

The momentum of a photon of energy 1 MeV in kg m/s, will be :

1. $0.33\times {10}^6$

2. $7\times {10}^{-24}$

3. ${10}^{-22}$

4. $5\times {10}^{-22}$

An electron is accelerated through a potential difference of 10,000 V. Its de-Broglie wavelength is, (nearly) : ($m_e= 9\times {10}^{-31}kg$)

1. 12.2 nm

2. $12.2\times {10}^{-13}m$

3. $12.2\times {10}^{-12}m$

4. $12.2\times {10}^{-14}m$

Light quanta with an energy of 4.9 eV eject photoelectrons from a light photosensitive surface with the work function $\left(\mathrm{\varphi }\right)$ = 4.5 eV. The maximum impulse that can be transmitted to the surface when each electron ejected, is

1.  3.45 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

2.  4.35 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

3.  5.35 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

4.  4.53 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

When monochromatic light of wavelength $\mathrm{\lambda }$ illuminates a metal surface then stopping potential for photoelectric current is 3${\mathrm{V}}_0$. If wavelength changes to 2$\mathrm{\lambda }$ then stopping potential becomes $V_0$. Stopping potential in case wavelength is changed to 3$\mathrm{\lambda }$, would be:

1.  $\frac{{\mathrm{V}}_0}{2}$

2.  $\frac{{\mathrm{V}}_0}{6}$

3.  $\frac{{\mathrm{V}}_0}{4}$

4.  $\frac{{\mathrm{V}}_0}{3}$

The stopping potential for photoelectrons emitted from a surface illuminated by the light of wavelength 400 nm is 500 mV. When the incident wavelength is changed to a new value, the stopping potential is found to be 800 mV. New wavelength is about:

1.  365 nm

2.  250 nm

3.  640 nm

4.  340 nm