Question

A monochromatic radiation of wavelength λ₁ is incident on a stationary atom as a result of which the wavelength of the photon after the collision becomes λ₂ and the recoiled atom has De Broglie’s wavelength λ₃. Then,

• A. λ₃ = $\sqrt{\lambda_{1}\mspace{6mu}\lambda_{2}}$
• B. λ₁ = $\frac{\lambda_{2}\mspace{6mu}\lambda_{3}}{\lambda_{2} + \lambda_{3}}$
• C. λ₁ =$\sqrt{\lambda_{1}^{2} + \lambda_{2}^{2}}$
• D. λ₃ = $\sqrt{\lambda_{1}^{2} + \lambda_{2}^{2}}$

Answer 1

Answer: (B)

λ₁ = $\frac{\lambda_{2}\mspace{6mu}\lambda_{3}}{\lambda_{2} + \lambda_{3}}$

Answer 2

Conservation of momentum yields

$\frac{h}{\lambda_{1}} + 0 = \frac{h}{\lambda_{2}} + mv$

⇒ $\frac{h}{\lambda_{1}} - \frac{h}{\lambda_{2}} = mv$

⇒ $\frac{1}{\lambda_{1}} - \mspace{6mu}\frac{1}{\lambda_{2}} = \frac{mv}{h}$

Since, $\frac{h}{mv}\mspace{6mu} = \mspace{6mu}\lambda_{3}$ ⇒$\frac{1}{\lambda_{1}} - \frac{1}{\lambda_{2}} = \frac{1}{\lambda_{3}}$

⇒ $\frac{1}{\lambda_{1}} = \frac{1}{\lambda_{2}} + \frac{1}{\lambda_{3}}$ ⇒ $\lambda_{1} = \frac{\lambda_{2}\lambda_{3}}{\lambda_{2} + \lambda_{3}}$