Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

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

An alpha particle moves along a circular path of radius $0.5 \ mm$ in a magnetic field of $2 \times 10^{-2} \ T$. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant $= 6.63 \times 10^{-34} \ J \ s$)

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(E)$ as