The wavelength of a microscopic particle of mass $9.1 \times 10^{-31} \ kg$ is $182 \ nm$. Its kinetic energy in $J$ is: $(h = 6.625 \times 10^{-34} \ J \ s)$

If $a_0$ is the radius of the first Bohr's orbit of the $H$-atom, the de-Broglie wavelength of an electron revolving in the second Bohr's orbit will be: (in $\pi a_0$)

Calculate the de-Broglie wavelength of an electron travelling at $1\%$ of the speed of light.

The de-Broglie wavelength of a particle of mass $6.63 \ g$ moving with a velocity of $100 \ ms^{-1}$ is

When light of wavelength $248 \, nm$ falls on a metal of threshold energy $3.0 \, eV$,the de-Broglie wavelength of emitted electrons is ............. $\mathring{A}$ (Round off to the Nearest Integer). [Use : $\sqrt{3}=1.73, h =6.63 \times 10^{-34} \, Js, m_e =9.1 \times 10^{-31} \, kg, c =3.0 \times 10^{8} \, ms^{-1}, 1 \, eV =1.6 \times 10^{-19} \, J$]

The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is
$(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$