JEE Advanced 2013 · Paper 1 · Chemistry Q40

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ChemistryInteger+4 / -1Question 40 of 60

The atomic masses of He and Ne are 4 and 20 a.m.u., 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

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Answer and worked solution

Official answer: 5

Thermal de Broglie wavelength $\lambda=\dfrac h{\sqrt{3mkT}}\propto\dfrac1{\sqrt{mT}}$ .

Ratio. He: $m=4$ , $T=200$ K; Ne: $m=20$ , $T=1000$ K:
$M=\sqrt{\frac{20\times1000}{4\times200}}=\sqrt{25}=5.$

Answer: 5

Topics:de Broglie and Heisenberg

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JEE Advanced 2013 · Paper 1, Chemistry question 40