If in a photoelectric experiment, the wavelength of incident radiation is reduced from \(6000~\mathring{A}\) to \(4000~\mathring{A}\), then:

1.	The stopping potential will decrease.
2.	The stopping potential will increase.
3.	The kinetic energy of emitted electrons will decrease.
4.	The value of the work function will decrease.

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is

(1) 2                     

(2) 4

(3) 6                     

(4) 10

A beam of light of wavelength \(\lambda\) and with illumination \(L\) falls on a clean surface of sodium. If \(N\) photoelectrons are emitted each with kinetic energy \(E\), then:
1. \(N \propto L \) and \(E \propto L \)
2.  \(N \propto L \) and \(E \propto \frac{1}{\lambda}\)
3.  \(N \propto \lambda\) and \(E \propto L \)
4.  \(N \propto \frac{1}{\lambda}\) and \(E \propto \frac{1}{L}\)

Light of frequency v is incident on a substance of threshold frequency ${\mathrm{v}}_0\left({\mathrm{v}}_0<\mathrm{v}\right)$. The energy of the emitted photo-electron will be 
(a) $\mathrm{h}\left(\mathrm{v}-{\mathrm{v}}_0\right)$                  (b) h/v
(c) $\mathrm{he}\left(\mathrm{v}-{\mathrm{v}}_0\right)$                (d) $\mathrm{h}/{\mathrm{v}}_0$

4 eV is the energy of the incident photon and the work function in 2eV. What is the stopping potential ?

(1) 2V                   

(2) 4V

(3) 6V                   

(4) $2\sqrt{2} \mathrm{V}$ 

When radiation of wavelength $\mathrm{\lambda }$ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is

(a) $2\mathrm{\lambda }$                 (b) $4\mathrm{\lambda }$
(c) $6\mathrm{\lambda }$                 (d) $8\mathrm{\lambda }$

If the energy of the photon is increased by a factor of 4, then its momentum 

(1) Does not change

(2) Decreases by a factor of 4

(3) Increases by a factor of 4

(4) Decreases by a factor of 2

In a photoemissive cell with executing wavelength $\mathrm{\lambda }$, the fastest electron has speed v. If the exciting wavelength is changed to 3$\mathrm{\lambda }$/4, the speed of the fastest emitted electron will be 
(a) $\mathrm{v}{\left(3/4\right)}^{1/2}$                              (b) $\mathrm{v}{\left(4/3\right)}^{1/2}$
(c) Less than $\mathrm{v}{\left(4/3\right)}^{1/2}$               (d) Greater than $\mathrm{v}{\left(4/3\right)}^{1/2}$

The figure shows the variation of photocurrent with anode potential for a photo-sensitive surface for three different radiations. Let ${\mathrm{I}}_{\mathrm{a}},{\mathrm{I}}_{\mathrm{b}}$ and ${\mathrm{I}}_{\mathrm{c}}$ be the intensities and ${\mathrm{f}}_{\mathrm{a}},{\mathrm{f}}_{\mathrm{b}}$ and ${\mathrm{f}}_{\mathrm{c}}$ be the frequencies for the curves a, b and c respectively. Then-

1. ${\mathrm{f}}_{\mathrm{a}}={\mathrm{f}}_{\mathrm{b}} \mathrm{and} {\mathrm{I}}_{\mathrm{a}}\ne {\mathrm{I}}_{\mathrm{b}}$ 
2. ${\mathrm{f}}_{\mathrm{a}}={\mathrm{f}}_{\mathrm{c}} \mathrm{and} {\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{c}}$
3. ${\mathrm{f}}_{\mathrm{a}}={\mathrm{f}}_{\mathrm{b}} \mathrm{and} {\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{c}}$
4. ${\mathrm{f}}_{\mathrm{a}}={\mathrm{f}}_{\mathrm{b}} \mathrm{and} {\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{b}}$

The stopping potential as a function of the frequency of the incident radiation is plotted for two different photoelectric surfaces \(A\) and \(B\). The graphs demonstrate that \(A\)'s work function is: