An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with electrons change?

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

$(a)$ For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10} \; m ?$
$(b)$ Also find the de Broglie wavelength of a neutron,in thermal equilibrium with matter,having an average kinetic energy of $\frac{3}{2} k T$ at $300 \; K$.

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

What is the de Broglie wavelength of
$(a)$ a bullet of mass $0.040 \; kg$ travelling at the speed of $1.0 \; km/s$
$(b)$ a ball of mass $0.060 \; kg$ moving at a speed of $1.0 \; m/s$,and
$(c)$ a dust particle of mass $1.0 \times 10^{-9} \; kg$ drifting with a speed of $2.2 \; m/s$?