The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

If a change in the momentum of an electron by $P_o$ results in a change of $0.25\%$ in its de-Broglie wavelength,then the initial momentum is:

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).