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

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

The de Broglie wavelength of a neutron at $27^{\circ}C$ is $\lambda$. What will be its de Broglie wavelength at $927^{\circ}C$?

$A$ light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then the ratio $\lambda : \lambda_1^2$ is ($h =$ Planck's constant,$c =$ velocity of light,$m =$ mass of electron).

If the momentum of an electron changes by $P$,then the de-Broglie wavelength associated with it changes by $5 \%$. The initial momentum of the electron is: (in $P$)