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 .............. $keV$.

The kinetic energy of an electron is $5 \ eV$. Calculate the de-Broglie wavelength associated with it in $\mathring{A}$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg)$

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

If the kinetic energies of an electron, an alpha particle, and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$, and $\varepsilon_3$ respectively, then:

The de Broglie wavelength of a molecule in a gas at room temperature $(300 \ K)$ is $\lambda_1$. If the temperature of the gas is increased to $600 \ K$,then the de Broglie wavelength of the same gas molecule becomes $..............$.