Question

The photo cut-off voltage in an experiment was found to be $1.5\;{\text{V}}$. The work function for the material used in the experiment was $4.22\;{\text{V}}$. The maximum kinetic energy for the photo electrons that was emitted was
A) 1.5 eV
B) 2.7 eV
C) 4.2 eV
D) 5.7 eV

Answer 1

The above numerical problem is related to the photoelectric effect. The photoelectric effect is the effect in which the electrons in an object start moving due to incident light on the object. The minimum energy of the light that initiates the movement of the electrons in the object is called the work function of the material. The kinetic energy of the emitted electrons is the same as the energy of the incident light.

Complete step by step answer:
Given: The photo cut-off voltage in the experiment is ${V_c} = 1.5\;{\text{V}}$, The work function of the material used in the experiment is $\phi = 4.22\;{\text{V}}$.
The expression to calculate the maximum kinetic of the emitted photoelectrons is given as:
${K_{\max }} = \phi - q{V_c}......\left( 1 \right)$
Here, q is the charge of the electron and it is the same as e.
Substitute $4.22\;{\text{V}}$for $\phi$ , 1 e for q and $1.5\;{\text{V}}$for ${V_c}$ in the expression (1) to calculate the maximum kinetic energy of the photo electrons.
$\begin{gathered} {K_{\max }} = 4.22\;{\text{eV}} - \left( {{\text{1}}\;{\text{e}}} \right)\left( {1.5\;{\text{V}}} \right) \\\ = 2.72\;{\text{eV}} \\\ \approx 2.7\;{\text{eV}} \\\ \end{gathered}$

Thus, the maximum kinetic energy for the photo electrons was $2.7\;{\text{eV}}$and the option (B) is the correct answer.

Additional Information: The work function of the material varies with the nature of material. If the work function of the material is less than the energy necessary for the emission of the photo electrons also become less.

Note: Always remember that the excess energy of the incident provides the kinetic energy to photo electrons for the movement. Substitute all the values in the expression in the same unit. Generally the work function of the material is given in the eV, so change the numerical value of incident light in eV, if it is given in Joule.