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

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

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

The de Broglie wavelength of a particle accelerated through a potential difference of $150 \ V$ is $10^{-10} \ m$. If it is accelerated through a potential difference of $600 \ V$,what will be its wavelength in $\mathring A$?

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ is: