Ncert Exercises & Solutions

A student performs an experiment on the photoelectric effect, using two materials A and B. A plot of $V_{s t o p} v e r s u s ν$ $V_{s t o p} v e r s u s ν$ is given in the figure.

(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron =1.6 $\times 10^{- 19}$ $\times 10^{- 19}$ C, find the value of h obtained from the experiment for both A and B.

Comment on whether it is consistent with Einstein's theory.

Hint: The work function of a material depends on the threshold frequency.

Step 1: Find the value of the work function for A and B.

(i) Given, threshold frequency of A is given by ${ν_{0}}_{A} = 5 \times 10^{14} H z a n d$ ${ν_{0}}_{A} = 5 \times 10^{14} H z a n d$

For B, $ν_{0_{B}} = 10 \times 10^{14} H z$ $ν_{0_{B}} = 10 \times 10^{14} H z$

We know that;

, $\begin{aligned} W o r k f u n c t i o n, ϕ & = h ν_{0} \\ \Rightarrow ϕ_{0} & \propto v_{0} \\ S o, \frac{ϕ_{0_{A}}}{ϕ_{_{0_{B}}}} & = \frac{5 \times 10^{14}}{10 \times 10^{14}} < 1 \\ \Rightarrow ϕ_{0_{A}} & < ϕ_{0_{B}} \end{aligned}$

Thus, work function of B is higher than that of A.

Step 2: Find the value of h from the experiment.

(ii)

$\begin{array}{ll} F o r m e t a l A, s l o p e = \frac{h}{e} & = \frac{2}{(10 - 5) \times 10^{14}} \\ h = \frac{2 e}{5 \times 10^{14}} = \frac{2 \times 1.6 \times 10^{- 19}}{5 \times 10^{14}} \\ = 6.4 \times 10^{- 34} Js \end{array}$
$\begin{array}{l} F o r m e t a l B, s l o p e = \frac{h}{e} = \frac{2.5}{(15 - 10) 10^{14}} \\ h = \frac{2.5 \times e}{5 \times 10^{14}} = \frac{2.5 \times 1.6 \times 10^{- 19}}{5 \times 10^{14}} \\ = 8 \times 10^{- 34} Js \end{array}$

Since, the value of h from experiment for metals A and B is different. Hence , experiment is not consistent with Einstein's theory.

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