An electron in stationary hydrogen atom jumps from $4^{\text {th }}$ energy level to ground level. The…
An electron in stationary hydrogen atom jumps from $4^{\text {th }}$ energy level to ground level. The velocity that the photon acquired as a result of electron transition will be ( $h=$ Planck's constant, $R=$ Rydberg's constant, $m=$ mass of photon)
$\frac{11 R h}{16 m}$
$\frac{15 R h}{16 m}$
$\frac{9 R h}{16 m}$
$\frac{13 R h}{16 m}$
Solution
The change in energy is given by
$\Delta E=h c R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here, energy of photon, $\Delta E=R h c\left(1-\frac{1}{4^2}\right)$
$\Rightarrow \Delta E=\frac{15}{16} R h c$
Now, energy of photon $=m c^2$
$\begin{aligned} & \Rightarrow m c^2=\frac{15}{16} R h c \\ & \Rightarrow c=\frac{15 R h}{16 m}\end{aligned}$
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