The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

(1) 540 nm                 

(2) 400 nm

(3) 310 nm                 

(4) 220 nm

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

Work function of a metal is 2.1 eV. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surface ?
(1) 4000 Å, 7500 Å                     

(2) 5500 Å, 6000 Å

(3) 4000 Å, 6000 Å                     

(4) None of these

The cathode of a photoelectric cell is changed such that the work function changes from ${\mathrm{W}}_1$ to ${\mathrm{W}}_2$ $\left({\mathrm{W}}_2>{\mathrm{W}}_1\right)$. If the current before and after the change are ${\mathrm{I}}_1$ and ${\mathrm{I}}_2$, all other conditions remaining unchanged, then (assuming $\mathrm{h\nu }>{\mathrm{W}}_2$) :

(1) ${\mathrm{I}}_1$= ${\mathrm{I}}_2$               

(2) ${\mathrm{I}}_1$< ${\mathrm{I}}_2$

(3) ${\mathrm{I}}_1$>${\mathrm{I}}_2$               

(4) ${\mathrm{I}}_1<{\mathrm{I}}_2<2{\mathrm{I}}_1$

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 \(E \propto \frac{1}{L}\)

Which of the following statements is correct

(1) The current in a photocell increases with increasing frequency of light

(2) The photocurrent is proportional to applied voltage

(3) The photocurrent increases with increasing intensity of light

(4) The stopping potential increases with increasing intensity of incident light

For the intensity \(I\) of a light of wavelength \(5000\) \(\mathring{A}\) the photoelectron saturation current is \(0.40\) $\mathrm{\mu A}$ and the stopping potential is \(1.36\) V, the work function of the metal is:
1. \(2.47\) eV
2. \(1.36\) eV
3. \(1.10\) eV
4. \(0.43\) eV

The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the surfaces of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B) 
(a) 1 : 1               (b) 1 : 2
(c) 1 : 3               (d) 1 : 4

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) $2\sqrt{2} \mathrm{V}$ 

The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is (taking h = $6\times {10}^{-34}$J-sec)

(a) 100                 (b) 1000
(c) $3\times {10}^{20}$           (d) $3\times {10}^{18}$