$A$ 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:

Assertion $(A):$ $A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of their de-Broglie wavelengths is unity.
Reason $(R):$ Here,we cannot apply the conservation of linear momentum.

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

The de Broglie wavelength of an electron in a metal at $27^{\circ}C$ is compared with the given mean distance between two electrons in the metal,which is $2 \times 10^{-10} \ m$. The ratio of the mean distance to the de Broglie wavelength is approximately:

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).

An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few $eV$. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c =$ speed of light in vacuum).