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Q.
Number of neutrons in $C^{12}$ and $ C^{14} \, are $
Odisha JEEOdisha JEE 2009Nuclei
Solution:
As, $ 4 \times 10^{3}=10^{20} \times h f $
$\Rightarrow f=\frac{4 \times 10^{3}}{10^{20} \times 6.023 \times 10^{-34}}$
$=6.64 \times 10^{16} Hz$
The obtained frequency lies in the band of $X$-rays.