A) \[h\nu \]
B) \[h\nu +eEd\]
C) \[h\nu -eEd\]
D) \[2h\nu -eEd\]
Correct Answer: C
Solution :
\[2h\nu \,\,=\,\,h\nu \,\,+\,\,K{{E}_{max}}\] \[K{{E}_{max}}=\,\,h\nu \] on reaching 2nd plates electron are retarded through potential difference eED so \[K{{E}_{remaining}}=\left( h\nu -eEd \right)\]You need to login to perform this action.
You will be redirected in
3 sec