The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

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 .................... $\mathring{A}$

Which of the following particles has the shortest de-Broglie wavelength,assuming they have the same kinetic energy?

$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

After absorbing a slowly moving neutron of mass $m_N$ (momentum $\approx 0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_1$ and $3m_1$ $(4m_1 = M + m_N)$,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: