Question

A student performs an experiment on photoelectric effect using two materials A and B. A plot of stopping potential $\left(V_0\right)$ vs frequency $\left(\upsilon \right)$ is as shown in the figure.The value of h obtained from the experiment for both A and B respectively is (Given electric charge of an electron = 1.6 $\times {10}^{-19}$ C)

• A. $3.2\times {10}^{-34}Js,4\times {10}^{-34}Js$
• B. $6.4\times {10}^{-34}Js,8\times {10}^{-34}Js$
• C. $1.2\times {10}^{-34}Js,3.2\times {10}^{-34}Js$
• D. $4.2\times {10}^{-34}Js,5\times {10}^{-34}Js$

Answer 1

Answer: (B)

$6.4\times {10}^{-34}Js,8\times {10}^{-34}Js$

$K.E=hv-\varphi$
for stoping potential $V_s=\frac{K.E}{e}$

$V_s=\frac{hv-\varphi }{e}$ ...........(1)

For metal A, Slope = $\frac{h}{e}=\frac{2-0}{(10-5)}\times {10}^{14}$

or h =$\frac{2\times e}{5\times {10}^{14}}=\frac{2\times 1.6\times {10}^{-19}}{5\times {10}^{14}}=6.4\times {10}^{-34}js$

For metal B , slope = $\frac{h}{e}=\frac{2.5-0}{(15-10}\times {10}^{14}$

h =$\frac{2.5\times e}{5\times {10}^{14}}=\frac{2.5\times 1.6\times {10}^{-19}}{5\times {10}^{14}}=8\times {10}^{-34}js$