Question

A particle of mass $m$ having de Broglie wavelength $\lambda$ collides elastically $(e=1)$ with another particle of mass $2m$ initialy at rest, de Broglie wavelength of particle of mass `2m` after collision is $\frac{n}{4}\lambda$, then `n` is [collision is head on ]

Answer 1

where $\lambda =\frac{h}{m{V}_0}\text{By Conservation of momentum}\Rightarrow m{V}_0=2m{V}_2+m{V}_1{V}_0=2V_2+V_1....\left(1\right)$
By, $e=1=\frac{V_2-V_1}{V_0}\Rightarrow V_0=V_2-V_1....\left(2\right)By\left(1\right)+\left(2\right)2V_0=3V_2\Rightarrow V_2=\frac{2}{3}{V}_0{\lambda }^{{}^{\prime }}=\frac{h}{2m{V}_2}=\frac{h\times 3}{2m2V_0}=\frac{3}{4}\frac{h}{m{V}_0}=\frac{3}{4}\lambda$