SATHEE: Modern Physics 1 Question 43

Modern Physics 1 Question 43

40. The recoil speed of a hydrogen atom after it emits a photon is going from $n = 5$ state to $n = 1$ state is $m / s . (1997, 1 M)$

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Solution:

From conservation of linear momentum,

|Momentum of recoil hydrogen atom $| = |$ Momentum of emitted photon|

or $m v = \frac{Δ E}{c}$

Here, $Δ E = E_{5} - E_{1} = - 13.6 \frac{1}{5^{2}} - \frac{1}{1^{2}} e V$

$\begin{aligned} = (13.6) (24 / 25) e V = 13.056 e V \\ = 13.056 \times 1.6 \times 10^{- 19} J = 2.09 \times 10^{- 18} J \end{aligned}$

and $m =$ mass of hydrogen atom $= 1.67 \times 10^{- 27} k g$

$\begin{aligned} ∴ v & = \frac{Δ E}{m c} = \frac{2.09 \times 10^{- 18}}{(1.67 \times 10^{- 27}) (3 \times 10^{8})}, \\ v & \approx 4.17 m / s \end{aligned}$

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Modern Physics 1 Question 43

40. The recoil speed of a hydrogen atom after it emits a photon is going from n=5 state to n=1 state is m/s.(1997,1M)

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40. The recoil speed of a hydrogen atom after it emits a photon is going from $n = 5$ state to $n = 1$ state is $m / s . (1997, 1 M)$