As collision is elastic, so laws of conservation of momentum and kinetic energy are obeyed.
According to law of conservation of momentum,
$$
\begin{aligned}
& \mathrm{m}_{\mathrm{A}} \mathrm{v}+\mathrm{m}_{\mathrm{B}}(0)=\mathrm{m}_{\mathrm{A}} \mathrm{v}_1+\mathrm{m}_{\mathrm{B}} \mathrm{v}_2 \\
\Rightarrow & \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}-\mathrm{v}_1\right)=\mathrm{m}_{\mathrm{B}} \mathrm{v}_2 \quad \ldots(\mathrm{i})
\end{aligned}
$$
According to law of conservation of kinetic energy
$$
\begin{aligned}
&\left(\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}^2+\frac{1}{2} \mathrm{~m}_{\mathrm{B}}(0)^2=\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_2^2\right) \\
&\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}^2=\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_2^2
\end{aligned}
$$

$$
\begin{aligned}
&\Rightarrow \quad \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}^2-\mathrm{v}_1^2\right)=\mathrm{m}_{\mathrm{B}} \mathrm{v}_2^2 \\
&\Rightarrow \quad \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}-\mathrm{v}_1\right)\left(\mathrm{v}+\mathrm{v}_1\right)=\mathrm{m}_{\mathrm{B}} \mathrm{v}_2^2
\end{aligned}
$$
Dividing Eq. (ii) by Eq. (i), we get, $v+v_1=v_2$ or $v=v_2-v_1$ Solving Eqs. (i) and (iii), we get,
$$
\begin{aligned}
&\left(m_A v-m_A v_1=m_B\left(v+v_1\right)=m_B v+m_B v_1\right) \\
&\left(m_A-m_B\right) v=v_1\left(m_A+m_B\right)
\end{aligned}
$$
So, $v_1=\left(\frac{m_A-m_B}{m_A+m_B}\right) v$ and $v_2=\left(\frac{2 m_A}{m_A+m_B}\right) v$
$$
\begin{aligned}
&\lambda_{\text {initial }}=\frac{h}{m_A v} \\
&\lambda_{\text {final }}=\frac{h}{m_A v_1}=\frac{h\left(m_A+m_B\right)}{m_A\left(m_A-m_B\right) v}
\end{aligned}
$$
Change in de-Broglie wavelength $(\Delta \lambda)$
$$
\begin{aligned}
&=\lambda_{\text {final }}-\lambda_{\text {initial }} \\
&=\frac{h}{m_A v}\left[\frac{m_A+m_B}{m_A-m_B}-1\right]=\frac{2 m_B h}{m_A\left(m_A-m_B\right) v}
\end{aligned}
$$