$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

$A$ particle of charge $q$ and mass $m$ enters a region of a transverse electric field of $E_{0} \hat{j}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de-Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths is . . . . . . .

Explain the particle-wave (dual) nature of matter.

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