When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is x 5 m s - 1 . Where x = _______. (Use: mass of hydrogen atom = 1 . 6 × 10 - 27 kg , charge of electron e = 1 . 6 × 10

Question

When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is x 5 m s - 1 . Where x = _______. (Use: mass of hydrogen atom = 1 . 6 × 10 - 27 kg , charge of electron e = 1 . 6 × 10 - 19 C )

TestHub · Accepted Answer

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

$\begin{array}{rcl} ΔE & = & - 3.4 eV - (- 13.6 eV) \\ = & 10.2 eV \end{array}$

The formula to calculate the recoil speed is given by

$v = \frac{∆ E}{m c} . . . (1)$

From equation (1), it follows that

$\begin{array}{rcl} v & = & \frac{10.2 eV \times \frac{1.6 \times 10^{- 19} J}{1 eV}}{1.6 \times 10^{- 27} kg \times 3 \times 10^{8} m s^{- 1}} \\ = & 3.4 m s^{- 1} \\ = & \frac{17}{5} m s^{- 1} \end{array}$

Therefore, $x = 17$ .

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