Question 61: Dual Nature of Matter and Radiation

Question

The de Broglie wavelength of an oxygen molecule at $27^{\circ} \mathrm{C}$ is $x 	imes 10^{-12} \mathrm{~m}$. The value of $x$ is (take Planck's constant $=6.63 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$, Boltzmann constant $=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{K}$, mass of oxygen molecule $=5.31 	imes 10^{-26} \mathrm{~kg}$ )

Accepted Answer

For a gas molecule, the average kinetic energy $(\mathrm{K})$ is given by kinetic theory:
$$ \mathrm{K}=\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{~T} $$
where $\mathrm{k}_{\mathrm{B}}$ is the Boltzmann constant and T is the absolute temperature.
The relationship between momentum ( p ) and kinetic energy ( K ) is:
$$ \mathrm{p}=\sqrt{2 \mathrm{mK}} $$
Substituting K:
$$ p=\sqrt{2 m\left(\frac{3}{2} k_B Tight)}=\sqrt{3{m k_B} \mathrm{~T}} \ldots (i)$$
The de Broglie wavelength is defined as:
$$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} \ldots $$
From both the relations,
$$ \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}} $$
The given temperature is $27^{\circ} \mathrm{C}$.
$$ \mathrm{T}=27+273=300 \mathrm{~K} $$
The given mass is $5.31 	imes 10^{-26} \mathrm{~kg}$.
So, the de Broglie wavelength is,
$$ \lambda=\frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes 5.31 	imes 10^{-26}}}=\frac{6.63 	imes 10^{-34}}{\sqrt{6595.02 	imes 10^{-49}}}=\frac{6.63 	imes 10^{-34}}{\sqrt{6.595 	imes 10^{-46}}}=\frac{6.63 	imes 10^{-34}}{2.568 	imes 10^{-23}} $$
$$\Rightarrow $$ $\lambda \approx 2.58 	imes 10^{-11} \mathrm{~m}=25.8 	imes 10^{-12} \mathrm{~m}$
$\Rightarrow \mathrm{x} 	imes 10^{-12} \mathrm{~m}=25.8 	imes 10^{-12} \mathrm{~m}$
$$ \Rightarrow x \approx 26 $$
Therefore, the value of x is 26 . Hence, the correct option is (D).

Answer

24

Answer

30

Answer

20

Answer

26

Dual Nature of Matter and Radiation

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