The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

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:

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

Both an electron and a photon have the same de-Broglie wavelength of $1.2 \ \text{Å}$. The ratio of their energies is nearly

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)