The kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

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

After absorbing a slowly moving neutron of mass $m_{N}$ (momentum $0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ $(6m_{1} = M + m_{N})$ 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 energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

What is the ratio of the de Broglie wavelengths of a proton and a neutron moving with the same velocity?

When the momentum of a proton is changed by an amount $p_0$,the corresponding change in the de-Broglie wavelength is found to be $0.25\%$. Then,the original momentum of the proton was