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

2 Views|Posted 6 months ago

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Alok Kumar Singh | Contributor-Level 10

6 months 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

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

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

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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: [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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