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CBSE Class 12 · 2017 · Delhi · Modern PhysicsShort Answer

Find the wavelength of the electron orbiting in the first excited state in a hydrogen atom.

Answer: $λ = π r_{2} \approx 6.65 \overset{˚}{A}$ .

Solution

First excited state:

n = 2

, orbit radius

r_{2} = 0.53 n^{2} = 0.53 \times 4 = 2.12 \overset{˚}{A}

.
The de-Broglie standing-wave condition gives

2 π r_{n} = nλ

:

λ = \frac{2 π r _{2}}{n} = \frac{2 π ( 2.12 )}{2} = π \times 2.12 \approx 6.66 \overset{˚}{A} .

λ \approx 6.65 \overset{˚}{A}

CBSE Class 12 Board · 2017 · Delhi · Subjective

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