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 =…

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

$\begin{array}{rcl}\mathrm{\Delta E}& =& -3.4 \mathrm{eV}-\left(-13.6 \mathrm{eV}\right)\\ & =& 10.2 \mathrm{eV}\end{array}$

The formula to calculate the recoil speed is given by

$v=\frac{∆E}{mc} ...\left(1\right)$

From equation (1), it follows that

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

Therefore, $x=17$.