The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. What will be its wavelength at $927^{\circ} C$?

Consider an electron $(m = 9.1 \times 10^{-31} \ kg)$ confined by electrical forces to move between two rigid walls separated by $1.0 \times 10^{-9} \ m$,which is about five atomic diameters. The quantized energy value for the lowest stationary state is

The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ($h=$ Planck's constant,$m=$ mass of electron).

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

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