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JEE Advanced 2016 Paper 2 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

JEE Advanced 2016 Paper 2 Physics: Paragraph 2

In a muonic hydrogen atom a negative muon (mass 207 me207\,m_e) is bound to a proton (mass 1836 me1836\,m_e) and the system is treated with the Bohr model, taking the finite proton mass into account through the reduced mass of the two-body system. For ordinary hydrogen (infinite nuclear mass) the first Bohr radius is 0.0529 nm0.0529\ \mathrm{nm} and the ground-state energy is −13.6 eV-13.6\ \mathrm{eV}. Both scale with the reduced mass μ\mu of the orbiting particle: the radius as 1/μ1/\mu and the energy magnitude as μ\mu. Use hc=1240 eV nmhc=1240\ \mathrm{eV\,nm}.

Question 1

Single correct option

The radius of the first Bohr orbit of the muon in this atom is closest to
  1. A. 256 fm256\ \mathrm{fm}
  2. B. 142 fm142\ \mathrm{fm}
  3. C. 512 fm512\ \mathrm{fm}
  4. D. 284 fm284\ \mathrm{fm}
Show answer and solution

Answer: (D) 284 fm284\ \mathrm{fm}

The reduced mass is μ=mμmpmμ+mp=207×1836207+1836 me=186.0 me\mu=\dfrac{m_\mu m_p}{m_\mu+m_p}=\dfrac{207\times1836}{207+1836}\,m_e=186.0\,m_e. The Bohr radius scales inversely with the reduced mass, so a=0.0529 nm186.0=2.84×10−4 nm=284 fma=\dfrac{0.0529\ \mathrm{nm}}{186.0}=2.84\times10^{-4}\ \mathrm{nm}=284\ \mathrm{fm}. If the proton mass is taken as infinite, μ=207 me\mu=207\,m_e and the radius is 0.0529/207=256 fm0.0529/207=256\ \mathrm{fm} (the value in option A), which is the typical mistake of ignoring the finite proton mass. This orbit radius is about 200 times smaller than that of ordinary hydrogen, which is why muonic atoms probe the nuclear size. The Bohr model is used here only to scale the hydrogen results.

Question 2

Single correct option

The wavelength of the photon emitted in the transition from n=2n=2 to n=1n=1 in this atom is closest to
  1. A. 0.65 nm0.65\ \mathrm{nm}
  2. B. 0.59 nm0.59\ \mathrm{nm}
  3. C. 0.49 nm0.49\ \mathrm{nm}
  4. D. 3.53 nm3.53\ \mathrm{nm}
Show answer and solution

Answer: (A) 0.65 nm0.65\ \mathrm{nm}

The ground-state energy magnitude is E1=13.6 eV×186.0=2530 eVE_1=13.6\ \mathrm{eV}\times186.0=2530\ \mathrm{eV}. The n=2→1n=2\to1 transition releases ΔE=E1(1−14)=0.75×2530=1897 eV\Delta E=E_1\left(1-\dfrac14\right)=0.75\times2530=1897\ \mathrm{eV}. The wavelength is λ=hcΔE=1240 eV nm1897 eV=0.654 nm\lambda=\dfrac{hc}{\Delta E}=\dfrac{1240\ \mathrm{eV\,nm}}{1897\ \mathrm{eV}}=0.654\ \mathrm{nm}, an X-ray photon. Using μ=207 me\mu=207\,m_e gives 0.59 nm0.59\ \mathrm{nm}, using the series limit E1E_1 alone gives 0.49 nm0.49\ \mathrm{nm}, and the 3→23\to2 line gives 3.53 nm3.53\ \mathrm{nm}. Only 0.65 nm0.65\ \mathrm{nm} is consistent with both the reduced mass and the correct transition. As a check, the photon energy is about 1.9 keV1.9\ \mathrm{keV}, which indeed lies in the soft X-ray range, as expected for muonic hydrogen.
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Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake