An $\alpha$ particle and a carbon $12$ atom have the same kinetic energy $K$. The ratio of their de-Broglie wavelengths $(\lambda_{\alpha} : \lambda_{C12})$ is

If the linear momentum of a proton is changed by $p_0$,then the de Broglie wavelength associated with the proton changes by $0.25 \%$. Then the initial linear momentum of the proton is

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 temperature of an ideal gas in $3$-dimensions is $300\, K$. The corresponding de-Broglie wavelength of the electron approximately at $300\, K$ is $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

$A$ particle of mass $4M$ at rest disintegrates into two particles of mass $M$ and $3M$ respectively,having non-zero velocities. The ratio of the de-Broglie wavelength of the particle of mass $M$ to that of mass $3M$ will be: