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=4N/C after traveling a distance d = 1m |
\[\therefore \,\,\,\frac{1}{2}mV_{\max }^{2}=eEd=4\,eV\] |
The energy of incident photon \[=\frac{1240}{200}\]\[=6.2\,\,eV\] |
From equation of photo electric effect |
\[\frac{1}{2}mv_{\max }^{2}=hv-{{\phi }_{0}}\] |
\[\therefore \,\,\,{{\phi }_{0}}=6.2-4=2.2\,\,eV\] |
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