$A$ proton and an electron are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_p$ denote the de-Broglie wavelengths of the electron and the proton,respectively.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

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

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following:

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would