JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 13)
An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 10^{-11} \mathrm{~m}$ while that for the electron in third excited state is $8.48 \times 10^{-10} \mathrm{~m}$. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is
4
3
9
16
Explanation
We know, $$\lambda = {h \over {mv}}$$
and $$mvr = {{nh} \over {2\pi }}$$
$$ \Rightarrow mv = {{nh} \over {2\pi r}}$$
So, $$\lambda = {h \over {nh}}2\pi r = 2\pi {r \over n}$$
$$ \Rightarrow \lambda \propto {r \over n}$$
We can write, $${{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{{r_g}} \over {{r_e}}}} \right)\left( {{{{n_e}} \over {{n_g}}}} \right)$$
$$ \Rightarrow {{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{5.3 \times {{10}^{ - 11}}} \over {84.8 \times {{10}^{ - 11}}}}} \right)\left( {{4 \over 1}} \right) = {1 \over 4}$$
$$ \Rightarrow {{{\lambda _e}} \over {{\lambda _g}}} = 4$$
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