At room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure,the de Broglie wavelength associated with a helium atom is compared with the average distance between two atoms. Which of the following relations is correct?

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is:

If a proton and an electron have the same linear momentum,then compared to the electron:

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.

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 $30.0 \; m/s$?