A) \[3.09\text{ }eV\]
B) \[1.41\text{ }eV\]
C) \[1.51\text{ }eV\]
D) \[1.68\text{ }eV.\]
Correct Answer: B
Solution :
\[\frac{hc}{\lambda }=\phi +{{(KE)}_{\max }}\] \[\Rightarrow \] \[\frac{1240}{400}=\phi +1.68\] \[\Rightarrow \] \[\phi =1.41\,eV\]You need to login to perform this action.
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