Dual Nature of Matter

JEE Main level, single correct, medium

46 of 200

Question 46 of 200

The temperature of an ideal gas in three dimensions is 300K.300K. The corresponding de-Broglie wavelength of the electron approximately at 300K300K is:
[me=\left[{m}_{e}=\right.mass of electron =9×10−31kg=9\times {10}^{-31}kg, h=h=Planck constant =6.6×10−34Js=6.6\times {10}^{-34}Js, kB={k}_{B}=Boltzmann constant=1.38×10−23JK−1]\left.=1.38\times {10}^{-23}J{K}^{-1}\right]

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Answer: (D) 6.26nm6.26nm

k=1.38×10−23JK−1k=1.38\times {10}^{-23}J{K}^{-1}
Mass of electron =9×10−31kg=9\times {10}^{-31}kg
⇒λD=h2m×32×kT\Rightarrow {\lambda }_{D}=\dfrac{h}{\sqrt{2m\times \dfrac{3}{2}\times kT}}=h3mkT=\dfrac{h}{\sqrt{3mkT}}
⇒λD=6.6×10−343×9×10−31×1.38×10−23×300=6.6×10−34105.72×10−27=62Aο\Rightarrow {\lambda }_{D}=\dfrac{6.6\times {10}^{-34}}{\sqrt{3\times 9\times {10}^{-31}\times 1.38\times {10}^{-23}\times 300}}=\dfrac{6.6\times {10}^{-34}}{105.72\times {10}^{-27}}=62A^{ο}
Thus x=62Aοx=62A^{ο}

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