If the linear momentum of a proton is changed by $p_0$,then the de Broglie wavelength associated with the proton changes by $0.25 \%$. Then the initial linear momentum of the proton is

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).

An electron of mass $m$ when accelerated through a potential difference $V$ has a de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be:

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

An electron of mass $m_{e}$ and a proton of mass $m_{p} = 1836 m_{e}$ are moving with the same speed. The ratio of their de Broglie wavelength $\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}}$ will be ....... .