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Physics · 2026 · 3 marks
CBSE 2026 · Region 2 · Set 2 · Q23
Calculate the de Broglie wavelength associated with an electron revolving in the second excited state of hydrogen atom. The ground state energy of…
Calculate the de Broglie wavelength associated with an electron revolving in the second excited state of hydrogen atom. The ground state energy of the hydrogen atom is -$\displaystyle 13.6$ eV. Take $\displaystyle \mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}$.s.
Marking-scheme solution
Total energy of the electron in the second excited state of the hydrogen atom: $\displaystyle \mathrm{E}_{n}=\dfrac{-13.6}{n^{2}} \mathrm{eV}$
$\displaystyle \mathrm{E}_{3}=\dfrac{-13.6}{3^{2}} \mathrm{eV}$
$\displaystyle \mathrm{E}_{3}=-1.51 \mathrm{eV}$
$\displaystyle \therefore$ Kinetic energy of the electron in this state, $\displaystyle \mathrm{K}=1.51 \mathrm{eV}$
de Broglie wavelength $\displaystyle \lambda=\dfrac{h}{\sqrt{2 m \mathrm{~K}}}$
$\displaystyle =\dfrac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.51 \times 1.6 \times 10^{-19}}}$
$\displaystyle \lambda \approx 10^{-9} \mathrm{~m}=1 \mathrm{~nm}$
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