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 $= 14.0076 \;u$)

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-rays 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 $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$

For an electron microscope,which of the following is false?

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .