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CBSE Class 12 · 2019 · Modern PhysicsShort Answer

Calculate the orbital period of the electron in the first excited state of a hydrogen atom.

Answer: $T \approx 1.2 \times 1 0^{- 15} s$ .

Solution

First excited state:

n = 2

.
Radius

r_{2} = 0.53 n^{2} \overset{˚}{A} = 0.53 \times 4 = 2.12 \overset{˚}{A} = 2.12 \times 1 0^{- 10} m

.
Speed

v_{2} = \frac{2.19 \times 1 0 ^{6}}{n} = \frac{2.19 \times 1 0 ^{6}}{2} = 1.095 \times 1 0^{6} m/s

.

T = \frac{2 π r _{2}}{v _{2}} = \frac{2 π ( 2.12 \times 1 0 ^{- 10} )}{1.095 \times 1 0 ^{6}} \approx 1.22 \times 1 0^{- 15} s .

T \approx 1.2 \times 1 0^{- 15} s

CBSE Class 12 Board · 2019 · Subjective

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