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 is 5.

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Detailed Solution

$\frac{λ_{H e}}{λ_{N e}} = \frac{M_{N e} V_{N e}}{M_{H e} V_{H e}} \Rightarrow \frac{\frac{M_{N e}}{M_{H e}} \sqrt{\frac{3 R T_{N e}}{M_{N e}}}}{M_{H e} \sqrt{\frac{3 R T_{H e}}{M_{H e}}}}$

$= \frac{M_{N e} \sqrt{\frac{T_{N e}}{M_{N e}}}}{M_{H e} \sqrt{\frac{T_{H e}}{M_{H e}}}} = \sqrt{\frac{M_{N e} T_{N e}}{M_{H e} T_{H e}}} \Rightarrow \sqrt{\frac{205 \times 1000}{4 \times 200}} \Rightarrow λ_{H e} = 5 λ_{N e}$