JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Modern Physics

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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 1 0^{- 26}

kg, Boltzman constant:

1.38 \times 1 0^{- 23}

J/K, Planck constant:

6.63 \times 1 0^{- 34}

Js)

(A) $0.24 \overset{˚}{A}$ ✓ Correct
(B) $0.20 \overset{˚}{A}$
(C) $0.34 \overset{˚}{A}$
(D) $0.44 \overset{˚}{A}$

Answer: (A)

Solution

The de Broglie wavelength is

λ = \frac{h}{m v}

, and at the rms speed

v_{r m s} = \frac{3 k T}{m}

this becomes

λ = \frac{h}{3 mk T}

. Putting in

h = 6.63 \times 1 0^{- 34}

m = 4.64 \times 1 0^{- 26}

k = 1.38 \times 1 0^{- 23}

and

T = 400 K

, the denominator is

3 (4.64 \times 1 0^{- 26}) (1.38 \times 1 0^{- 23}) (400) \approx 2.77 \times 1 0^{- 23}

, so

λ \approx 2.4 \times 1 0^{- 11} m \approx 0.24 \overset{˚}{A}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Modern Physics

Solution

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