Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

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 a moving bus with speed $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now, when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal,then the ratio of velocities of the proton and the $\alpha$-particle will be

If the de Broglie wavelength of a particle is equal to the wavelength of a photon,then the energy of the photon is .....

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