The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of the proton,$\lambda_2$ is the wavelength associated with the photon,and the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

If a proton and an electron have the same linear momentum,then compared to the electron:

An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .