A) 0.43 eV
B) 0.55 eV
C) 1.10 eV
D) 1.53 eV
Correct Answer: C
Solution :
[c]: Energy of a photon\[E=\frac{hc}{\lambda }\]or\[E\propto \frac{1}{\lambda }\] \[\therefore \]\[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] or\[{{E}_{2}}={{E}_{1}}\times \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=1.23\times \frac{10000}{5000}=2.46eV\] According to Einstein?s photoelectric equation \[h\upsilon -{{\phi }_{0}}=\frac{1}{2}m{{v}^{2}}_{\max }=e{{V}_{s}}\] or\[{{\phi }_{0}}=h{{v}_{2}}-e{{V}_{s}}={{E}_{2}}-e{{V}_{s}}\]
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