The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.

$A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles,$\lambda_1 / \lambda_2$ is

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (kelvin) and mass $m$ is:

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