An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential, so that the wavelength is doubled (in $V$)?

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

The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

What will be the ratio of de-Broglie wavelengths of a proton and an $\alpha$-particle of the same energy?