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

The de Broglie wavelength for an electron accelerated through a potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is . . . . . . .

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).

What is the kinetic energy of a neutron associated with a de Broglie wavelength of $1.40 \times 10^{-10} \ m$? (Mass of neutron $m = 1.675 \times 10^{-27} \ kg$,$h = 6.6 \times 10^{-34} \ J \cdot s$).