An electron microscope works on the principle of......

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

What voltage should be applied to an electron microscope to produce electrons with a wavelength of $\lambda = 1.0 \ \mathring{A}$ (in $V$)?

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through the same potential of $100\, V$. What should nearly be the ratio of their wavelengths? $(m_{P} = 1.00727\, u, m_{e} = 0.00055\, u)$

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to