$A$ 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 $2 : 1$)

The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?

$A$ proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$.

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

What is the de Broglie wavelength associated with
$(a)$ an electron moving with a speed of $5.4 \times 10^{6} \; m/s$,and
$(b)$ a ball of mass $150 \; g$ travelling at $30.0 \; m/s$?

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: