The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

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).

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$. The rate of change of de Broglie wavelength associated with the electron varies inversely as the $n^{\text{th}}$ power of time. The numerical value of $n$ is . . . . . . .

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.