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Dmitriy789 [7]
3 years ago
11

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

Chemistry
1 answer:
andrey2020 [161]3 years ago
3 0

<u>Answer:</u> The de-Broglie's wavelength of a hydrogen molecule is 3.26\AA

<u>Explanation:</u>

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

E=\frac{3}{2}kT

where,

E = kinetic energy of the particles  = ?

k = Boltzmann constant  = 1.38\times 10^{-23}J/K

T = temperature of the particle = 30 K

Putting values in above equation, we get:

E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J

  • Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used:  1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or 2\times 10^{-3}kg  

According to mole concept:

6.022\times 10^{23} number of molecules occupy 1 mole of a gas.

As, 6.022\times 10^{23} number of hydrogen molecules has a mass of 2\times 10^{-3}kg

So, 1 molecule of hydrogen will have a mass of = \frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg

  • To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

\lambda = De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of 1 hydrogen molecule = 3.32\times 10^{-27}kg

E_k = kinetic energy of the particle = 6.21\times 10^{-22}J

Putting values in above equation, we get:

\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}

\lambda=3.26\times 10^{-10}m=3.26\AA    (Conversion factor: 1\AA=10^{-10}m )

Hence, the de-Broglie's wavelength of a hydrogen molecule is 3.26\AA

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