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The atomic masses of He and Ne are 4 and 20 amu , respectively. The value of the de Broglie wavelength of He gas at - 73 ° C is "M" times that of the de Broglie wavelength of Ne at 727 ° C . M is:

Correct answer

0

Step-by-step solution

DeBroglie wave length is given as λ = h 2 m K.E = h 2 m 3 2 k b · T = h 3 k b m T k b is Boltzmann's constant λ He = h 3 k b 2 0 0 4 × 1.66 × 1 0 - 2 7 T = 200 K m He = 4 × 1.66 × 1 0 - 2 7 kgs λ Ne = h 3 k b 1 0 0 0 2 0 × 1.66 × 1 0 - 2 7 T = 1000 K m Ne = 2 0 × 1.66 × 1 0 - 2 7 kgs λ He λ Ne = 1 0 00 × 2 0 2 0 0 × 4 = 2 5 = 5

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