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Question: Assume that a molecule is moving with the root mean square speed at temperature 300k.The de Broglie ...

Assume that a molecule is moving with the root mean square speed at temperature 300k.The de Broglie wavelength of nitrogen molecule is : (Atomic mass of nitrogen=14,0076u, h=66.3×10-34 Js

kB=1.38×10−23JK−1,1u=1.66×10−27kgk_{B} = 1.38 \times 10^{- 23}JK^{- 1},1u = 1.66 \times 10^{- 27}kg):

A

2.25×10−11m2.25 \times 10^{- 11}m

B

2.25×10−11m2.25 \times 10^{- 11}m

C

2.25×10−11m2.25 \times 10^{- 11}m

D

2.25×10−11m2.25 \times 10^{- 11}m

Answer

2.25×10−11m2.25 \times 10^{- 11}m

Explanation

Solution

: Mean kinetic energy of a molecule

=12mv2rms=32kBT= \frac{1}{2}m{v^{2}}_{rms} = \frac{3}{2}k_{B}T

∴vrms=3kBTm\therefore v_{rms} = \sqrt{\frac{3k_{B}T}{m}}

Here m=2×14.0076=28.02um = 2 \times 14.0076 = 28.02u

As

}{= \frac{6.63 \times 10^{- 34}}{\sqrt{3 \times (28.02 \times 1.66 \times 10^{- 27}) \times 1.38 \times 10^{- 23} \times 300}}}$$ $$= 2.75 \times 10^{- 11}m$$