If the momentum of a proton is changed by $p_0$,the corresponding change in the de Broglie wavelength is $0.25 \%$. The initial momentum of the proton is:

An electron (mass $m$) with an initial velocity $\overrightarrow{v} = v_{0} \hat{i} \left(v_{0} > 0\right)$ is moving in an electric field $\overrightarrow{E} = -E_{0} \hat{i} \left(E_{0} > 0\right)$ where $E_{0}$ is constant. If at $t = 0$ the de Broglie wavelength is $\lambda_{0} = \frac{h}{mv_{0}}$,then its de Broglie wavelength after time $t$ is given by:

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

Find the typical de Broglie wavelength associated with a $He$ atom in helium gas at room temperature $(27^{\circ}C)$ and $1 \; atm$ pressure; and compare it with the mean separation between two atoms under these conditions.

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)