JEE knocking the door!

Chemistry Level 2

The atomic masses of Helium and Neon are 4 and 20 amu respectively. What is the ratio of de Broglie wavelength of He gas at $-73^\circ C$ compared to that of Ne gas at $727^\circ C$ ?

The answer is of the form $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers. Enter your answer as $a+b$ .

The answer is 6.

1 solution

Aditya Kumar
May 18, 2016

De-broglie wavelength is given by: $\lambda=\frac{h}{mv}$ . Here $h$ is plank's constant, $m$ is atomic mass, $v$ is rms velocity.

$v_{rms}$ is given by: $v_{rms}=\sqrt{\frac{3RT}{M}}$

Now on inserting the values and taking the ratio,we get: $\frac{{\lambda}_{He}}{{\lambda}_{Ne}}=\frac{5}{1}$ .

Here, $a=5,b=1$ . Hence, $\boxed{a+b=6}$ .

JEE knocking the door!

The answer is 6.

1 solution

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