Question

An electron of stationary hydrogen atom jumps from $4th$ 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)

• A. $\frac{9Rh}{16m}$
• B. $\frac{11hR}{16m}$
• C. $\frac{13hR}{16m}$
• D. $\frac{15hR}{16m}$

Answer 1

The correct option is D $\frac{15hR}{16m}$

The change in energy is given by

$\Delta E=Rhc(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Here, energy of photon, $\Delta E=Rhc(1-\frac{1}{4^2})$

$⟹\Delta E=\frac{15}{16}Rhc$

Now, energy of photon=$m{c}^2$

$⟹m{c}^2=\frac{15}{16}Rhc$

$⟹c=\frac{15Rh}{16m}$

Hence, answer is option-(D).