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 $0.5 \ MeV$ ).

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:

An electron,an $\alpha$-particle,and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

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

An electron with an initial kinetic energy of $100 \, eV$ is accelerated through a potential difference of $50 \, V$. The final de-Broglie wavelength of the electron becomes .................... $\mathring{A}$

The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is: