With what potential an electron should be accelerated so that its de Broglie wavelength becomes equal to the wavelength of the first line of the Lyman series for the $He^+$ ion?

The de Broglie wavelength for a deuteron can be given by:

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

Assuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

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 $(\lambda_1/\lambda_2)^2$.