If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

$A$ proton and an $\alpha$-particle are accelerated from rest by $2\,V$ and $4\,V$ potentials, respectively. The ratio of their de-Broglie wavelength is: (in $:1$)

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
$(A) \ n^{-3/2} \ (B) \ n^{-1} \ (C) \ n^{1/2} \ (D) \ n$

$A$ particle having electric charge $3 \times 10^{-19} \text{ C}$ and mass $6 \times 10^{-27} \text{ kg}$ is accelerated by applying an electric potential of $1.21 \text{ V}$. The wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12} \text{ m}$. The value of $\alpha$ is . . . . . . . (Take Planck's constant $h = 6.6 \times 10^{-34} \text{ J} \cdot \text{s}$)