An electron is accelerated through a potential difference of $10,000 \ V$. Its de-Broglie wavelength is nearly: $(m_{e} = 9 \times 10^{-31} \ kg)$

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (kelvin) and mass $m$ is:

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is: