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
Solution
<p>We know, $\lambda = {h \over {mv}}$</p>
<p>and $mvr = {{nh} \over {2\pi }}$</p>
<p>$\Rightarrow mv = {{nh} \over {2\pi r}}$</p>
<p>So, $\lambda = {h \over {nh}}2\pi r = 2\pi {r \over n}$</p>
<p>$\Rightarrow \lambda \propto {r \over n}$</p>
<p>We can write, $${{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{{r_g}} \over {{r_e}}}} \right)\left( {{{{n_e}} \over {{n_g}}}} \right)$$</p>
<p>$$ \Rightarrow {{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{5.3 \times {{10}^{ - 11}}} \over {84.8 \times {{10}^{ - 11}}}}} \right)\left( {{4 \over 1}} \right) = {1 \over 4}$$</p>
<p>$\Rightarrow {{{\lambda _e}} \over {{\lambda _g}}} = 4$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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