A) 1.0V greater
B) 1.0V smaller
C) 0.5V greater
D) 0.5V smaller
E) 0. IV greater
Correct Answer: A
Solution :
\[E=\frac{hc}{\lambda }\Rightarrow E\propto \frac{1}{\lambda }\] \[\Rightarrow \] \[\frac{E}{E}=\frac{400}{300}=1.33\] But\[E=e{{V}_{s}},\text{ }{{V}_{s}}\]being stopping potential. Thus, stopping potential for photoelectrons from a surface becomes approximately 1.0 V greater.You need to login to perform this action.
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