The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

Which of the following has the largest de Broglie wavelength?

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

Assuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)

The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)

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