An electron is moving with initial velocity $\vec{v} = V_0\hat{j}$ in a magnetic field $\vec{B} = B_0\hat{i}$. Its de-Broglie wavelength will

The ratio of de-Broglie wavelength of an $\alpha$-particle and proton accelerated from rest by the same potential is $\dfrac{1}{\sqrt{m}}$. The value of $m$ is

The proton and $\alpha$ - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$ - particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton)

Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass m are $V_1$ and $V_2$ respectively, then ($h$ = Planck's constant)