When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is 5 x m s − 1 . Where x = _______. (Use: mass of hydrogen atom = 1.6 × 1 0 − 27 k g , charge of electron e = 1.6 × 1

Question

When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is 5 x ​ m s − 1 . Where x = _______. (Use: mass of hydrogen atom = 1.6 × 1 0 − 27 k g , charge of electron e = 1.6 × 1 0 − 19 C )

TestHub · Accepted Answer

According to the above diagram, the difference in energy between the two states is given by

$Δ E = - 3.4 e V - (- 13.6 e V) = 10.2 e V$ The formula to calculate the recoil speed is given by

$v = \frac{Δ E}{m c} ... (1)$ From equation (1), it follows that

$v = \frac{10.2 e V \times \frac{1.6 \times 1 0 ^{- 19} J}{1 e V}}{1.6 \times 1 0 ^{- 27} k g \times 3 \times 1 0 ^{8} m s ^{- 1}} = 3.4 m s^{- 1} = \frac{17}{5} m s^{- 1}$

Therefore, $x = 17$ .

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