If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

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:

The de-Broglie wavelength $\lambda$ of a particle is:

If the kinetic energy of a free electron is doubled,its de Broglie wavelength becomes ........... times.

An electron microscope uses which property of the electron?

An electron in an electron microscope with initial velocity $v_{0} \hat{i}$ enters a region of a stray transverse electric field $E_{0} \hat{j}$. The time taken for the change in its de-Broglie wavelength from the initial value of $\lambda$ to $\lambda / 3$ is proportional to