Ques 18:

18. What is the de Broglie wavelength of a nitrogen molecule in air of 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of Nitrogen = 14.0075 u)

Open in App

Solution

Dear Student,
$T = 300 K, M = 28 V_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 * 8.314 * 300}{28}} = 16.35 m / s . M_{(i n k g)} V_{r m s} = \frac{h}{λ} λ = \frac{h}{M V_{r m s}} = \frac{6.634 * 10^{- 34}}{28 * 1.66 * 10^{- 27} * 16.35} = 8.73 * 10^{- 10} m R e g a r d s$

Suggest Corrections

Similar questions

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Q. Assume that a molecule is moving with the root mean square speed at temperature 300 K. The de Broglie wavelength of nitrogen molecule is (Atomic mass of nitrogen = 14.0076 u, h =

6.63 \times 10^{- 27} J s, k_{B} = 1.38 \times 10^{- 23} J K^{- 1}, 1 u = 1.66 . \times 10^{- 27} k g)

Q. 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

)

Q. Assertion :The de Broglie wavelength of a molecule varies inversely as the square root of temperature. Reason: The root mean square velocity of the molecule depends on the temperature.

Q. Assertion : The de-Broglie wavelength of a molecule varies inversely as the square root of temperature.

Reason : The root mean square velocity of the molecule depends on the temperature.

Join BYJU'S Learning Program

Ques 18: 18. What is the de Broglie wavelength of a nitrogen molecule in air of 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of Nitrogen = 14.0075 u)

Dear Student, T=300 K ,M=28Vrms=3RTM=3*8.314*30028=16.35 m/s .Min kgVrms=hλλ=hMVrms=6.634*10-3428*1.66*10-27*16.35=8.73*10-10 m Regards

Ques 18:

18. What is the de Broglie wavelength of a nitrogen molecule in air of 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of Nitrogen = 14.0075 u)

Dear Student,
$T = 300 K, M = 28 V_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 * 8.314 * 300}{28}} = 16.35 m / s . M_{(i n k g)} V_{r m s} = \frac{h}{λ} λ = \frac{h}{M V_{r m s}} = \frac{6.634 * 10^{- 34}}{28 * 1.66 * 10^{- 27} * 16.35} = 8.73 * 10^{- 10} m R e g a r d s$