What is the de Broglie wavelength of a nitrogen molecule in air at $300 \;K$ (in $nm$)? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $= 14.0076 \;u$)

$A$ particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 \ V$. Another particle $B$ of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500 \ V$. The ratio of de-Broglie wavelength $\frac{\lambda_A}{\lambda_B}$ is close to

The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$

An electron is moving with initial velocity $\vec{v} = V_0 \hat{j}$ in a magnetic field $\vec{B} = B_0 \hat{i}$. Its de-Broglie wavelength will

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is