Wave Optics

Question

If the average life time of an excited state of hydrogen is of the order of 10^–8 s, estimate how many rotation an electron makes when it is in the state n = 2 and before it suffers a transition to state n = 1. Bohr radius = 5.3 × 10^–11 m.

Solution

Average lifetime of an excited state of hydrogen = 10^-8 sec
Bohr radius, r = 5.3

\times

10^-11 m

Velocity of electron in the nth orbit of hydrogen atom,

v_{n} = \frac{v_{1}}{n} = \frac{2.19 \times 10^{6}}{n} m / s

If,

n = 2, v_{n} = \frac{2.19 \times 10^{6}}{n} m / s

Radius of n = 2 orbit,

r_{n} = n^{2} r_{1} = 4 \times Bohr radius

\Rightarrow

r_{n} = 4 \times 5.3 \times 10^{- 11} m

Number of revolutions made in 1 sec

= \frac{v_{n}}{2 πr} = \frac{2.19 \times 10^{6}}{2 \times 2 π \times 4 \times 5.3 \times 10^{- 11}}

Therefore, number of revolutions made in 10^-8 sec,

= \frac{2.19 \times 10^{6} \times 10^{- 8}}{2 \times 2 π \times 4 \times 5.3 \times 10^{- 11}} = 8.22 \times 10^{6} revolutions .

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