Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wave length of nitrogen molecule is close to: (Given: nitrogen molecule weight: $$4.64 \times 10^{-26}\,\text{kg}$$, Boltzman constant: $$1.38 \times 10^{-23}\,\text{J K}^{-1}$$, Planck constant: $$6.63 \times 10^{-34}\,\text{J s}$$)

We have to determine the de-Broglie wavelength of a nitrogen molecule that is moving with its root-mean-square (r.m.s.) speed at a temperature of $$400\ \text{K}$$.

First we recall the expression for the r.m.s. speed of a gas molecule in an ideal gas:

$$v_{\mathrm{rms}}=\sqrt{\dfrac{3k_{\mathrm{B}}T}{m}}$$

Here $$k_{\mathrm{B}}=1.38\times10^{-23}\,\text{J\,K}^{-1}$$ is the Boltzmann constant, $$T=400\ \text{K}$$ is the absolute temperature, and $$m=4.64\times10^{-26}\,\text{kg}$$ is the mass of one nitrogen molecule. Substituting the given values we get

$$v_{\mathrm{rms}}=\sqrt{\dfrac{3\,(1.38\times10^{-23})\,(400)}{4.64\times10^{-26}}}$$

We first evaluate the numerator inside the square root:

$$3\,(1.38\times10^{-23})\,(400)=3\,(552\times10^{-23})=1656\times10^{-23}=1.656\times10^{-20}$$

Now we form the complete fraction:

$$\dfrac{1.656\times10^{-20}}{4.64\times10^{-26}} =1.656\times10^{-20}\times\dfrac{1}{4.64\times10^{-26}} =\dfrac{1.656}{4.64}\times10^{6} \approx0.3576\times10^{6}=3.576\times10^{5}$$

Taking the square root of the result gives

$$v_{\mathrm{rms}}=\sqrt{3.576\times10^{5}}\ \text{m s}^{-1}\approx5.98\times10^{2}\ \text{m s}^{-1}$$

Next, we recall the de-Broglie relation which links a particle’s momentum to its wavelength:

$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$

where $$h=6.63\times10^{-34}\,\text{J\,s}$$ is Planck’s constant, $$m=4.64\times10^{-26}\,\text{kg}$$ is again the mass and $$v=v_{\mathrm{rms}}$$ is the speed just calculated. Substituting the numbers, we obtain

$$\lambda=\dfrac{6.63\times10^{-34}}{(4.64\times10^{-26})(5.98\times10^{2})}$$

We multiply the denominator term-by-term:

$$(4.64\times10^{-26})(5.98\times10^{2}) =4.64\times5.98\times10^{-26+2} =27.77\times10^{-24} =2.777\times10^{-23}$$

Hence

$$\lambda=\dfrac{6.63\times10^{-34}}{2.777\times10^{-23}} =\dfrac{6.63}{2.777}\times10^{-34+23} \approx2.387\times10^{-11}\ \text{m}$$

Finally, we convert metres to ångströms using $$1\ \text{\AA}=10^{-10}\ \text{m}:$$

$$\lambda=2.387\times10^{-11}\ \text{m} =\dfrac{2.387\times10^{-11}}{10^{-10}}\ \text{\AA} =0.2387\ \text{\AA}\approx0.24\ \text{\AA}$$

Hence, the correct answer is Option A.