$A$ deuteron is accelerated through $500 \ V$. For the same de Broglie wavelength,a singly ionized helium ion is accelerated through a potential difference of $V \ V$. Find the value of $V$.

$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is

If an electron and a proton have the same de-Broglie wavelength,then the kinetic energy of the electron is

$A$ particle of mass $9.1 \times 10^{-31} \, \text{kg}$ travels in a medium with a speed of $10^{6} \, \text{m/s}$ and a photon of radiation with linear momentum $10^{-27} \, \text{kg} \cdot \text{m/s}$ travels in vacuum. The wavelength of the photon is $....$ times the wavelength of the particle.

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

$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be: