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

$A$ proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is $2000$ times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(\lambda_{p})$ and that of the electron $(\lambda_{e})$?

In an electron microscope,which nature of the electron is used?

$A$ particle having electric charge $3 \times 10^{-19} \text{ C}$ and mass $6 \times 10^{-27} \text{ kg}$ is accelerated by applying an electric potential of $1.21 \text{ V}$. The wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12} \text{ m}$. The value of $\alpha$ is . . . . . . . (Take Planck's constant $h = 6.6 \times 10^{-34} \text{ J} \cdot \text{s}$)

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.