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Structure of Atom question

2013 · Shift 1 · Q1
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Structure of Atom question

2013 · Shift 1 · Q1

JEE AdvancedChemistryStructure of AtomNumerical+4 / −1
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 –73oC is “M” times that of the de Broglie wavelength of Ne at 727oC. M is
Numerical answer
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Correct answer: 5

  1. For a gas particle, the de Broglie wavelength is
λ=hp\lambda = \frac{h}{p}λ=ph​

For thermal motion, momentum is related to temperature by

p∝mTp \propto \sqrt{mT}p∝mT​

so

λ∝1mT\lambda \propto \frac{1}{\sqrt{mT}}λ∝mT​1​
  1. Therefore, for He and Ne,
λHeλNe=mNeTNemHeTHe\frac{\lambda_{\mathrm{He}}}{\lambda_{\mathrm{Ne}}} = \sqrt{\frac{m_{\mathrm{Ne}}T_{\mathrm{Ne}}}{m_{\mathrm{He}}T_{\mathrm{He}}}}λNe​λHe​​=mHe​THe​mNe​TNe​​​
  1. Convert temperatures to Kelvin:
  • He at −73∘C-73^\circ \mathrm{C}−73∘C:
THe=273−73=200 KT_{\mathrm{He}} = 273 - 73 = 200\,\mathrm{K}THe​=273−73=200K
  • Ne at 727∘C727^\circ \mathrm{C}727∘C:
TNe=273+727=1000 KT_{\mathrm{Ne}} = 273 + 727 = 1000\,\mathrm{K}TNe​=273+727=1000K
  1. Substitute masses and temperatures:
λHeλNe=20×10004×200\frac{\lambda_{\mathrm{He}}}{\lambda_{\mathrm{Ne}}} = \sqrt{\frac{20 \times 1000}{4 \times 200}}λNe​λHe​​=4×20020×1000​​ =20000800=25=5= \sqrt{\frac{20000}{800}} = \sqrt{25} = 5=80020000​​=25​=5
  1. Hence,
λHe=5 λNe\lambda_{\mathrm{He}} = 5\,\lambda_{\mathrm{Ne}}λHe​=5λNe​

So the required value is

M=5M = 5M=5
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