A) \[4\,eV\]
B) \[6.2\,eV\]
C) \[2\,eV\]
D) \[2.2\,eV\]
Correct Answer: D
Solution :
The electron ejected with maximum speed \[{{v}_{\max }}\] are stopped by electric field \[E\text{ }=4N/C\]after travelling a distance \[d=1m\] \[\frac{1}{2}mv_{\max }^{2}=eEd=4eV\] The energy of incident photon \[=\frac{1240}{200}=6.2eV\] From equation of photo electric effect \[\frac{1}{2}mv_{\max }^{2}=hv-{{\phi }_{0}}\] \[\therefore \,\,{{\phi }_{0}}=6.2-4=2.2eV\]
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