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JEE Main 2017 · Physics · Atoms, Nuclei and Dual Nature of Radiation · hard

A beam of electrons, each of energy 12.5 eV, is passed through hydrogen gas in its ground state, and the atoms excited by collisions are allowed to…

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A beam of electrons, each of energy 12.5 eV12.5\ \mathrm{eV}, is passed through hydrogen gas in its ground state, and the atoms excited by collisions are allowed to de-excite. Taking hc=1240 eV nmhc = 1240\ \mathrm{eV\,nm} and En=−13.6/n2 eVE_n = -13.6/n^2\ \mathrm{eV}, the number of different spectral lines emitted and the shortest wavelength among them are respectively
  1. A. 33 lines, 103 nm103\ \mathrm{nm}
  2. B. 33 lines, 656 nm656\ \mathrm{nm}
  3. C. 66 lines, 103 nm103\ \mathrm{nm}
  4. D. 66 lines, 97 nm97\ \mathrm{nm}
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Answer: (A) 33 lines, 103 nm103\ \mathrm{nm}

A colliding electron can give an atom any amount of energy up to its own, provided it matches a level. Excitation energies from n=1n = 1: to n=2n=2, 10.2 eV10.2\ \mathrm{eV}; to n=3n=3, 12.09 eV12.09\ \mathrm{eV}; to n=4n=4, 12.75 eV12.75\ \mathrm{eV}. Since 12.5 eV12.5\ \mathrm{eV} lies between the last two, the highest level reached is n=3n = 3. De-excitation from n=3n=3 gives lines 3→23\to2, 3→13\to1, 2→12\to1: 33 lines. The shortest wavelength is for 3→13\to1: λ=124012.09≈103 nm\lambda = \dfrac{1240}{12.09} \approx 103\ \mathrm{nm}. Counting n=4n=4 would give 66 lines and 97 nm97\ \mathrm{nm}, but that level is not accessible.
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