JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 16)
A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 $$\times$$ 10$$^{-15}$$ eVs) :
99.3 nm
94.1 nm
974 nm
941 nm
Explanation
$\frac{h c}{\lambda}=+13.6 \mathrm{eV}\left[\frac{1}{1}-\frac{1}{4^{2}}\right]$
$$ \Rightarrow \frac{4 \times 10^{-15} \times 3 \times 10^{-8}}{\lambda}=13.6\left[\frac{15}{16}\right] $$
$\lambda=94.1 \mathrm{~nm}$
$$ \Rightarrow \frac{4 \times 10^{-15} \times 3 \times 10^{-8}}{\lambda}=13.6\left[\frac{15}{16}\right] $$
$\lambda=94.1 \mathrm{~nm}$
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