$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal,then the ratio of velocities of the proton and the $\alpha$-particle will be

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

If a photon,an electron,and a uranium nucleus have the same de Broglie wavelength,which one will have the highest energy?

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).