Question 63 of 200

Particle $A$ of mass ${\text{m}}_{\text{A}}=\dfrac{\text{m}}{2}$ moving along the $\text{x}$ -axis with velocity ${\text{v}}_{0}$ collides elastically with another particle $\text{B}$ at rest having mass ${\text{m}}_{\text{B}}=\dfrac{\text{m}}{3}$. If both the particles move along the $\text{x}$ -axis after the collision, the change $∆\lambda$ in the wavelength of the particle $\text{A}$, in terms of its de-Broglie wavelength $\left({\lambda }_{0}\right)$ before the collision is:

A$∆\lambda =\dfrac{3}{2}{\lambda }_{0}$B$∆\lambda =\dfrac{5}{2}{\lambda }_{0}$C$∆\lambda =2{\lambda }_{0}$D$∆\lambda =4{\lambda }_{0}$

Show answer and solution▾

Answer: (D) $∆\lambda =4{\lambda }_{0}$

${v}_{1}=\dfrac{2\left(m/3\right)0}{\left(\dfrac{m}{2}+\dfrac{m}{3}\right)}+\dfrac{\left(\dfrac{m}{2}-\dfrac{m}{3}\right)v}{\left(\dfrac{m}{2}+\dfrac{m}{3}\right)}=\dfrac{v}{5}$
For particle $A$,
Initial de-Broglie wavelength
${\lambda }_{0}=\dfrac{h}{\dfrac{m}{2}v}=\dfrac{2h}{mv}$
Final de-Broglie wavelength after collision.
${\lambda }_{1}=\dfrac{h}{\dfrac{m}{2}\dfrac{v}{5}}=\dfrac{10h}{mv}=5{\lambda }_{0}$
Change in De-Broglie wavelength $\Delta \lambda ={\lambda }_{i}-{\lambda }_{0}=4{\lambda }_{0}$