The graph shows the variation of de Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$,where '$V$' is the accelerating potential for four particles $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?

Find the de Broglie wavelength associated with a Helium $(He)$ atom in Helium gas at room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure.

Consider two particles of different masses. In which of the following situations will the heavier of the two particles have a smaller de-Broglie wavelength?

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

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: