Two particles of equal masses are moving with equal speeds at an angle $60^o$. The de-Broglie wavelength of these particles is $\lambda$. Find the de-Broglie wavelength of the particles in the frame of the centre of mass of the particles.

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

The de-Broglie wavelength of a neutron at $27^oC$ is $\lambda$. What will be its wavelength at $927^oC$?

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 $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following:

If an electron has an energy such that its de-Broglie wavelength is $5500 \ \text{Å}$,then the energy value of that electron is $(h = 6.6 \times 10^{-34} \ \text{Js}, m_e = 9.1 \times 10^{-31} \ \text{kg})$.