In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} \ m$, the minimum electron energy required is close to .............. $keV$.

If the momentum of an electron changes by $P$,then the de-Broglie wavelength associated with it changes by $5 \%$. The initial momentum of the electron is: (in $P$)

The ratio of the accelerating potentials required to accelerate $(i)$ an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is (mass of $\alpha$-particle = $6.4 \times 10^{-27} \text{ kg}$, mass of proton = $1.6 \times 10^{-27} \text{ kg}$)

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

After absorbing a slowly moving neutron of mass $m_{N}$ (momentum $0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ $(6m_{1} = M + m_{N})$ respectively. If the de-Broglie wavelength of the nucleus with mass $m_{1}$ is $\lambda$,then the de-Broglie wavelength of the other nucleus will be: