A) 2 V
B) less than 2 V
C) zero
D) more than 2 V
Correct Answer: D
Solution :
\[KE+\phi =E\] \[\phi =E-KE\] For same matter, \[\phi \] remains constant. \[\therefore \] \[{{E}_{1}}-{{(KE)}_{1}}={{E}_{2}}-{{(KE)}_{2}}=\] constant \[=\frac{hc}{\lambda }-eV=\]constant As\[\lambda \]is decreased therefore stopping potential will increase.
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