If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

The de-Broglie wavelength of a neutron at $27^oC$ is $\lambda$. What will be its wavelength at $927^oC$?

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

The graph shows the variation of de Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$,where '$V$' is the accelerating potential for four particles $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

If the kinetic energies of an electron,an alpha particle,and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$,and $\varepsilon_3$ respectively,then: