If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

Which of the following figures represents the variation of particle momentum $p$ and the associated de-Broglie wavelength $\lambda$?

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \ V$. After this,their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$ is

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is

$A$ proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is