JEE PYQ: Atomic Physics Question 35
Question 35 - 2020 (06 Sep Shift 2)
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to:
(Given: nitrogen molecule weight: $4.64 \times 10^{-26}$ kg, Boltzman constant: $1.38 \times 10^{-23}$ J/K, Planck constant: $6.63 \times 10^{-34}$ J.s)
(1) 0.24 A
(2) 0.20 A
(3) 0.34 A
(4) 0.44 A
Show Answer
Answer: (1)
Solution
Rms speed of gas molecule, $V_{rms} = \sqrt{\frac{3kT}{m}}$. de Broglie wavelength, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mk}} = \frac{h}{\sqrt{2m \times \frac{1}{2}mV_{rms}^2}} = \frac{h}{\sqrt{m \times \frac{3}{2}kT}} = \frac{h}{\sqrt{3mkT}}$. Substituting: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 4.64 \times 10^{-26} \times 1.38 \times 10^{-23} \times 400}} = 0.24$ A.
BETA
