$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of $6600 \text{ Å}$ is nearly (Planck's constant $= 6.6 \times 10^{-34} \text{ J s}$ and mass of electron $= 9 \times 10^{-31} \text{ kg}$)

When a particle is restricted to move along the $x$-axis between $x=0$ and $x=a$,where $a$ is of nanometer dimension,its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region correspond to the formation of standing waves with nodes at its ends $x=0$ and $x=a$. The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = \frac{p^2}{2m}$. Thus,the energy of the particle can be denoted by a quantum number $n$ taking values $1, 2, 3, \ldots$ ($n=1$,called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line $x=0$ to $x=a$. Take $h = 6.6 \times 10^{-34} \ J \ s$ and $e = 1.6 \times 10^{-19} \ C$.
$1.$ The allowed energy for the particle for a particular value of $n$ is proportional to
$(A) \ a^{-2} \ (B) \ a^{-3/2} \ (C) \ a^{-1} \ (D) \ a^2$
$2.$ If the mass of the particle is $m = 1.0 \times 10^{-30} \ kg$ and $a = 6.6 \ \text{nm}$,the energy of the particle in its ground state is closest to
$(A) \ 0.8 \ \text{meV} \ (B) \ 8 \ \text{meV} \ (C) \ 80 \ \text{meV} \ (D) \ 800 \ \text{meV}$
$3.$ The speed of the particle,that can take discrete values,is proportional to
$(A) \ n^{-3/2} \ (B) \ n^{-1} \ (C) \ n^{1/2} \ (D) \ n$

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

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to: