A) \[5\times {{10}^{16}}Hz\]
B) \[5\times {{10}^{17}}Hz\]
C) \[0.5\times {{10}^{18}}Hz\]
D) \[5\times {{10}^{18}}Hz\]
E) \[5\times {{10}^{-16}}Hz\]
Correct Answer: D
Solution :
\[\because \] \[E=m{{c}^{2}};E=hv\] \[\therefore \] \[m{{c}^{2}}=hv\] \[v=\frac{m{{c}^{2}}}{h}=\frac{1.1\times {{10}^{-23}}\times {{(3\times {{10}^{8}})}^{2}}}{6.6\times {{10}^{-34}}}\] \[=5\times {{10}^{18}}Hz\]
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