If the kinetic energy of a particle is increased to $16$ times, the percentage change in the de Broglie wavelength of a particle is (in $\%$)

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

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 ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

The kinetic energy of an electron is tripled,then the de-Broglie wavelength associated with it will change by a factor of: