A) \[\frac{8}{3}eV\]
B) \[8eV\]
C) \[\frac{10}{3}eV\]
D) \[\frac{20}{3}eV\]
Correct Answer: B
Solution :
[b] \[hv=\phi +{{K}_{\max }}\] \[hv=2+4=6eV\] \[\text{Now, }h\left( \frac{5}{3}v \right)\text{ =2+}{{\text{K}}_{\max }}\] \[{{K}_{\max }}=\frac{5}{3}\times 6-2\Rightarrow {{K}_{\max }}=8eV.\]
You need to login to perform this action.
You will be redirected in
3 sec
