An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

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.

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is