Question 114 of 239

When a hydrogen atom going from $n=2$ to $n=1$ emits a photon, its recoil speed is $\dfrac{x}{5}m{s}^{-1}$. Where $x=$_______. (Use: mass of hydrogen atom $=1.6\times {10}^{-27}kg$, charge of electron $e=1.6\times {10}^{-19}C)$

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Answer: 17

According to the above diagram, the difference in energy between the two states is given by
$\begin{matrix}\Delta E & = & -3.4eV-\left(-13.6eV\right) \\ & = & 10.2eV\end{matrix}$
The formula to calculate the recoil speed is given by
$v=\dfrac{∆E}{mc}...\left(1\right)$
From equation (1), it follows that
$\begin{matrix}v & = & \dfrac{10.2eV\times \dfrac{1.6\times {10}^{-19}J}{1eV}}{1.6\times {10}^{-27}kg\times 3\times {10}^{8}m{s}^{-1}} \\ & = & 3.4m{s}^{-1} \\ & = & \dfrac{17}{5}m{s}^{-1}\end{matrix}$
Therefore, $x=17$.