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Question
Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy of the hydrogen atom is −13.6 eV.
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Solution
Given: Ground state energy = −13.6 eV
For a hydrogen atom,
En = `(-13.6)/n^2 eV`
First excited state ⇒ n = 2
E2 = `(-13.6)/2^2`
= `(-13.6)/4`
= −3.4 eV
In the Bohr model:
K.E. = ∣En∣ = 3.4 eV
= 3.4 × 1.6 × 10−19
= 5.44 × 10−19 J
By using de-Broglie relation:
λ = `h/p`
Since,
K.E.= `p^2/(2 m)`
p = `sqrt(2 m K.E.)`
Thus,
λ = `h/sqrt(2 m K.E.)`
= `(6.63 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 5.44 xx 10^-19))`
= 6.66 × 10−10 m
= 6.66 Å
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[Take, h = 6.6 × 10-34 J-s]
