Question

For intensity of a light of wavelength 5000 $\buildrel _\circ \over {\mathrm{A}}$ the photoelectron saturation current is 0.40$\mu$A and stopping potential is 1.36V, the work function of metal is:
(A) 2.47 eV
(B) 1.36 eV
(C) 1.10 eV
(D) 0.43 eV

Answer 1

The term ‘work function’ in the photoelectric effect describes the minimum energy which is required to bring a free electron from the interior of the metal to the surface. Stopping potential is the negative potential (given to anode-cathode) which causes the flow of current to stop.
Formula used:
If V is the stopping potential for light of frequency incident on the cathode, the relation for stopping potential is:
$\dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} = eV$

Complete answer:
Upon giving the light of given wavelength, a part of energy is used in bringing electrons to the surface, remaining is used in giving it energy to take part in the process of conduction. The relation of wavelength and energy is given as:
$E = \dfrac{{hc}}{\lambda }$.
The threshold or the work function is therefore similarly written as:
${\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}$.
We are given the stopping potential. When such potential V is applied in reverse polarity, it causes the electron to stop or we can say that it stops the current in the circuit. The work function is therefore obtained as:

& \dfrac{{hc}}{\lambda } - eV = {\phi _0} \cr & \Rightarrow \dfrac{{12400eV \buildrel _\circ \over {\mathrm{A}}}}{{5000 \buildrel _\circ \over {\mathrm{A}}}} - 1.36eV = {\phi _0} \cr & \Rightarrow (2.48 - 1.36)eV = 1.12eV \cr} $$ Where hc has been directly substituted as $12400eV \buildrel _\circ \over {\mathrm{A}}$ which serves as a good approximate value and eases the calculations. **Therefore, from the given set of options, the correct option is (C) 1.10 eV.** **Note:** We say that an electron that is present in a region of potential 1 V attains an energy of 1 eV. Electron- volts is therefore a unit of energy. We are given potential in volts. Here energy is given per electron, then it will be V volts. So, the energy acquired by any electron is simply ‘V’ electron volts.