The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 × 10^-³¹kg

h = Planck constant = 6.6 × 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 × 10^-²³ JK^-¹]

Option 1 -

3.25 nm

Option 2 -

6.26 nm

Option 3 -

2.26 nm

Option 4 -

8.46 nm

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Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

De Broglie wavelength

$λ = \frac{h}{m V} = \frac{h}{\sqrt[]{2 m E}} E \to K . E .$

$E = \frac{3}{2}$ KT for gas

$S o λ = \frac{h}{\sqrt[]{3 m K T}} = \frac{6.6 \times 1 0^{- 34}}{\sqrt[]{3 \times 9 \times 1 0^{- 31} \times 1.38 \times 1 0^{- 23} \times 300}}$

$\begin{array}{l} λ = 6.26 \times 1 0^{- 9} m \\ λ = 6.26 n m \end{array}$

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The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 × 10^-³¹kg

h = Planck constant = 6.6 × 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 × 10^-²³ JK^-¹]

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The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is: [me = mass of electron = 9 × 10-31kg h = Planck constant = 6.6 × 10-34 Js kB = Boltzmann constant = 1.38 × 10-23 JK-1]

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The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 × 10^-³¹kg

h = Planck constant = 6.6 × 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 × 10^-²³ JK^-¹]