An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 1 0 − 11 m while that for the electron in third excited state is 8.48 × 1 0 − 10 m . The ratio of the de Broglie wavelengths of electron in the excited state to that in ground state is :

λ = nh 2 π r h so with r ∝ n 2 , λ ∝ n λ 4 λ 1 = r 4 /4 r 1 /1 = 8.48 × 1 0 − 10 5.3 × 1 0 − 11 × 4 = 4 1 Answer (A)

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PhysicsModern PhysicsDe Broglie & Matter WavesEasy2 minQB

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An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 1 0^{- 11} m$ while that for the electron in third excited state is $8.48 \times 1 0^{- 10} m$ . The ratio of the de Broglie wavelengths of electron in the excited state to that in ground state is :

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Solution:

$λ = \frac{2 π r h}{nh}$ so with $r \propto n^{2}$ , $λ \propto n$

$\frac{λ _{1}}{λ _{4}} = \frac{r _{1} /1}{r _{4} /4} = \frac{5.3 \times 1 0 ^{- 11} \times 4}{8.48 \times 1 0 ^{- 10}} = \frac{1}{4}$

Answer (A)

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: QB

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