Two particles A₁ and A₂ of masses m₁, m₂ (m₁ > m₂) have the same de-Broglie wavelength. Then,

de-Broglie wavelength $\lambda =\frac{h}{mv}$  where, mv = p (moment)
$\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$
Here, h is a constant.
So,         $p\propto \frac{1}{\lambda }\Rightarrow \frac{p_1}{p_2}=\frac{{\lambda }_2}{{\lambda }_1}$

But        $\left({\lambda }_1={\lambda }_2\right)=\lambda$

Then,    $\frac{p_1}{p_2}=\frac{\lambda }{\lambda }=1\Rightarrow p_1=p_2$ 

Thus, their momenta is same. 

Also,     $E=\frac{1}{2}\cdot m{v}^2=\frac{1}{2}\frac{m{v}^2\times m}{m}$

                $=\frac{1}{2}\frac{m^2{v}^2}{m}=\frac{1}{2}\frac{p^2}{m}$

Here, p is constant  $E\propto \frac{1}{m}$

$$\begin{gathered} \therefore \frac{E_1}{E_2}=\frac{m_2}{m_1}<1 \\ \Rightarrow E_1<E_2 \end{gathered}$$