Question

An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass $m$ acquired a result of photon emission will be :

($R$ is Rydberg constant and $h$ is Planck's constant)

• A. $\frac{25m}{24hR}$
• B. $\frac{24m}{25hR}$
• C. $\frac{24hR}{25m}$
• D. $\frac{25hR}{24m}$

Answer 1

The correct option is C $\frac{24hR}{25m}$

We know that , $\frac{1}{\lambda }=R(1/n_1^2-1/n_2^2)$

Here, $n_1=1,n_2=5$ so $\frac{1}{\lambda }=R(1/1^2-1/5^2)=\left(24/25\right)R$

Now, photon energy $E=\frac{hc}{\lambda }=\left(24/25\right)hcR$

As atom of mass $m$ acquired a velocity as a result of photo emission so using momentum conservation ,

the momentum photon = momentum of atom= $p$

Thus, $p=E/c=\left(24/25\right)hR$

Velocity of atom, $v=p/m=\frac{24hR}{25m}$