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Dual Nature of Radiation and Matter Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Dual Nature of Radiation and Matter

11.19 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)

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Payal Gupta | Contributor-Level 10

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11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $\times$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $\times 10^{- 27}$ kg

So, m = 28.0152 $\times$ 1.66 $\times 10^{- 27}$ kg = 4.65 $\times 10^{- 26}$ kg

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Boltzmann constant, k = 1.38 $\times 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2} k T)$ of the nitrogen molecule with root mean square speed ( $v_{r m s}$ ) as:

$\frac{1}{2} m {v_{r m s}}^{2}$ = $\frac{3}{2} k T$

$v_{r m s} = \sqrt{\frac{3 k T}{m}}$ = $\sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 300}{4.65 \times 10^{- 26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$λ = \frac{h}{m v_{r m s}}$ = $\frac{6.626 \times 10^{- 34}}{4.65 \times 10^{- 26} \times 516.814}$ = 2.76 $\times 10^{- 11}$ m = 0.0276 nm

Ther

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11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $\times$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $\times 10^{- 27}$ kg

So, m = 28.0152 $\times$ 1.66 $\times 10^{- 27}$ kg = 4.65 $\times 10^{- 26}$ kg

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Boltzmann constant, k = 1.38 $\times 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2} k T)$ of the nitrogen molecule with root mean square speed ( $v_{r m s}$ ) as:

$\frac{1}{2} m {v_{r m s}}^{2}$ = $\frac{3}{2} k T$

$v_{r m s} = \sqrt{\frac{3 k T}{m}}$ = $\sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 300}{4.65 \times 10^{- 26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$λ = \frac{h}{m v_{r m s}}$ = $\frac{6.626 \times 10^{- 34}}{4.65 \times 10^{- 26} \times 516.814}$ = 2.76 $\times 10^{- 11}$ m = 0.0276 nm

Therefore, De Broglie wavelength is 0.0276 nm.

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11.19 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)

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