Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

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: $c = 3 \times 10^8 \ m/s$)

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

$A$ proton and an electron are accelerated by the same potential difference. If their de-Broglie wavelengths are $\lambda_p$ and $\lambda_e$ respectively,then:

The figure shows four situations in which an electron is moving in an electric or magnetic field. In which case is the de Broglie wavelength of the electron increasing?

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