Physics - Modern Physics Question with Solution | TestHub

1. Balmer Series Wavelength:

$$\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$$

where $R_H$ is the Rydberg constant and $n>2$ is an integer.

2. Longest Wavelength in Balmer Series (n=3):

$$\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R_H\left(\frac{1}{4}-\frac{1}{9}\right)=R_H\left(\frac{9-4}{36}\right)=\frac{5}{36}{R}_H$$

Therefore, $\lambda =\frac{36}{5R_H}$

3. Photon Momentum:

$$p=\frac{h}{\lambda }$$

4. Momentum Conservation:

$$p_{atom}+p_{photon}=0$$

$$p_{atom}=-p_{photon}$$

5. Recoil Velocity:

$$v=\frac{p_{atom}}{M}$$

Steps:

1. Calculate the momentum of the photon emitted during the transition corresponding to the longest wavelength of the Balmer series (n=3 to n=2).

$$\frac{1}{\lambda }=\frac{5}{36}{R}_H$$

$$p=\frac{h}{\lambda }=\frac{5}{36}h{R}_H$$

2. Apply the principle of conservation of momentum. Since the initial momentum of the hydrogen atom is zero, the magnitude of the momentum of the recoiling atom is equal to the momentum of the emitted photon.

$$p_{atom}=p_{photon}=\frac{5}{36}h{R}_H$$

3. Calculate the recoil velocity of the hydrogen atom.

$$v=\frac{p_{atom}}{M}=\frac{5h{R}_H}{36M}=\frac{5}{36}\frac{h{R}_H}{M}$$

Since $R_H=\frac{R_{\infty }}{c}$, and $R_{\infty }$ is the Rydberg constant, we can write the recoil velocity as:

$$v=\frac{5}{36}\frac{Rh}{M}$$

Final Answer:

The recoil velocity acquired by the atom is $\frac{5}{36}\frac{Rh}{M}$.