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 travelling a distance\[~d=1\,m\] |
\[\therefore \] \[\frac{1}{2}\,m\,{{V}_{\max }}^{2}=eE\,\,d=4e\,V\] |
The energy of incident photon\[=\frac{1240}{200}=6.2\,eV\] From equation of photo electric effect\[\frac{1}{2}m{{v}_{\max }}^{2}=h\upsilon -{{\phi }_{0}}\] |
\[\therefore \] \[{{\phi }_{0}}=6.2-4=2.2\,eV.\] |
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