Question

Calculate the stopping potential $V_{s2}$

• A. 4.47
• B. 3.16
• C. 2.76
• D. 5.28

Answer 1

Answer: (A)

4.47

Using the Einstein's equation for photoelectric effect:

In first case, $e{V}_{s1}=\frac{hc}{{\lambda }_1}-W$ ....(1) where $W=$ work function of metal surface.

And in second case, $e{V}_{s2}=\frac{hc}{{\lambda }_2}-W$ ....(2)

From (1) and (2), $e{V}_{s2}=\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}+e{V}_{s1}=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{190\times {10}^{-9}}-\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{550\times {10}^{-9}}+(1.6\times {10}^{-19})\left(0.19\right)$

$=7.152\times {10}^{-19}J$

or $V_{s2}=\frac{7.152\times {10}^{-19}}{1.6\times {10}^{-19}}=4.47V$