The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

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

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.

If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be:

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.