A) \[{{10}^{19}}\,\,Hz\]
B) \[{{10}^{18}}\,\,\,Hz\]
C) \[{{10}^{16}}\,\,Hz\]
D) \[{{10}^{20}}\,\,Hz\] \[(1\,eV=1.6\times {{10}^{-19}}J\] and \[h=6.63\times {{10}^{-34}}J-\sec )\]
Correct Answer: A
Solution :
\[h{{\nu }_{o}}=eV\]\[\therefore \,\,\,{{\nu }_{o}}=\frac{eV}{h}=\frac{1.6\times {{10}^{-19}}\times 42000}{6.63\times {{10}^{-34}}}={{10}^{19}}\,Hz\]
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