The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was

The ratio of the accelerating potentials required to accelerate $(i)$ an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is (mass of $\alpha$-particle = $6.4 \times 10^{-27} \text{ kg}$, mass of proton = $1.6 \times 10^{-27} \text{ kg}$)

An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$. The rate of change of de Broglie wavelength associated with the electron varies inversely as the $n^{\text{th}}$ power of time. The numerical value of $n$ is . . . . . . .

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

$(a)$ 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 $300 \; K$.

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is