If the kinetic energy of a free electron doubles,its de-Broglie wavelength $\lambda$ changes by a factor of:

$A$ 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

$A$ particle of mass $8 \mu g$ in motion collides with another stationary particle of mass $4 \mu g$. If the collision is perfectly elastic and one-dimensional, the ratio of their de Broglie wavelengths after collision is (in $2 : 1$)

When the momentum of a proton is changed by an amount $p_0$,the corresponding change in the de-Broglie wavelength is found to be $0.25\%$. Then,the original momentum of the proton was

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

An electron beam is incident on a crystal. The interplanar spacing of the crystal is $b$. Which of the following de Broglie wavelengths of the electron beam will result in direct reflection?