The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

$A$ 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

If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

$A$ charged particle $q_0$ of mass $m_0$ is projected along the $y-$axis at $t = 0$ from the origin with a velocity $v_0$. If a uniform electric field $E_0$ also exists along the $x-$axis,then the time at which the de Broglie wavelength of the particle becomes half of its initial value is:

The idea of matter waves was given by

Find the ratio of the de Broglie wavelength of an oxygen molecule at $21^oC$ to that of a nitrogen molecule at $63^oC$.