Question 150: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

Given, Planck's constant, h = 6.6 $$	imes$$ 10
$$-$$34
Js
Boltzmann constant, k
B
= 1.38 $$	imes$$ 10
$$-$$23
J/K
Mass of an electron, m
e
= 9 $$	imes$$ 10
$$-$$31
kg
Temperature of an ideal gas, T = 300 K
As we know that, de-Broglie wavelength,
$$\lambda  = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$ .... (i)
Here, E is the kinetic energy,
$$E = {{3{K_B}T} \over 2}$$
Substituting value of E in Eq. (i), we get
$$\lambda  = {h \over {\sqrt {3m{K_B}T} }}$$
Substituting the given values in the above equation, we get
$$\lambda  = {{6.6 	imes {{10}^{ - 34}}} \over {\sqrt {3 	imes 9 	imes {{10}^{ - 31}} 	imes 1.38 	imes {{10}^{ - 23}} 	imes 300} }}$$
= 6.26 nm
$$	herefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

Answer

6.26 nm

Answer

8.46 nm

Answer

2.26 nm

Answer

3.25 nm

Dual Nature of Matter and Radiation

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

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