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Atoms and Nuclei question

2008 · Shift 1 · Q65
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  5. /2008 · Shift 1 · Q65

Atoms and Nuclei question

2008 · Shift 1 · Q65

JEE AdvancedPhysicsAtoms and NucleiMCQ+3 / −1
In a mixture of H - He +^++ gas (He +^++ is singly ionized He atom), H atoms and He +^++ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He +^++ ions (by collisions). Assume that the Bohr model of atom is exactly valid.The wavelength of light emitted in the visible region by He +^++ ions after collisions with H atoms is
  1. A
    6.5×10−76.5\times10^{-7}6.5×10−7 m
  2. B
    5.6×10−75.6\times10^{-7}5.6×10−7 m
  3. C
    4.8×10−74.8\times10^{-7}4.8×10−7 m
  4. D
    4.0×10−74.0\times10^{-7}4.0×10−7 m
View written solutionFree

Correct answer: C

  1. Initial excitation states

    • Hydrogen atom HHH is excited to its first excited state: n=2n=2n=2
    • Helium ion He+He^+He+ is hydrogen-like with nuclear charge Z=2Z=2Z=2, and it is also excited to its first excited state: n=2n=2n=2

  2. Energy levels using Bohr model

    For a hydrogen-like species, En=−13.6Z2n2 eVE_n=-\frac{13.6 Z^2}{n^2}\text{ eV}En​=−n213.6Z2​ eV

    For hydrogen (Z=1)(Z=1)(Z=1)

    E1=−13.6 eV,E2=−13.64=−3.4 eVE_1=-13.6\text{ eV},\qquad E_2=-\frac{13.6}{4}=-3.4\text{ eV}E1​=−13.6 eV,E2​=−413.6​=−3.4 eV So the excitation energy of HHH in the first excited state is ΔEH=E2−E1=(−3.4)−(−13.6)=10.2 eV\Delta E_H=E_2-E_1=(-3.4)-(-13.6)=10.2\text{ eV}ΔEH​=E2​−E1​=(−3.4)−(−13.6)=10.2 eV

  3. Energy levels of He+He^+He+

    For He+He^+He+, Z=2Z=2Z=2: En=−13.6×4n2=−54.4n2 eVE_n=-\frac{13.6\times 4}{n^2}=-\frac{54.4}{n^2}\text{ eV}En​=−n213.6×4​=−n254.4​ eV

    Thus, E1=−54.4 eV,E2=−13.6 eVE_1=-54.4\text{ eV},\qquad E_2=-13.6\text{ eV}E1​=−54.4 eV,E2​=−13.6 eV

    Since He+He^+He+ is initially in n=2n=2n=2, its energy is −13.6 -13.6\,−13.6eV.

  4. Collision: hydrogen transfers its total excitation energy to He+He^+He+

    Hydrogen gives its excitation energy: 10.2 eV10.2\text{ eV}10.2 eV

    So He+He^+He+ absorbs this energy while starting from n=2n=2n=2: Efinal=−13.6+10.2=−3.4 eVE_{\text{final}}=-13.6+10.2=-3.4\text{ eV}Efinal​=−13.6+10.2=−3.4 eV

    Now find the corresponding level of He+He^+He+: −54.4n2=−3.4-\frac{54.4}{n^2}=-3.4−n254.4​=−3.4 n2=54.43.4=16n^2=\frac{54.4}{3.4}=16n2=3.454.4​=16 n=4n=4n=4

    So after collision, He+He^+He+ goes from n=2→n=4n=2 \to n=4n=2→n=4

  5. Visible photon emitted by He+He^+He+ after collision

    Now the ion can de-excite from n=4n=4n=4.

    Possible downward transitions:

    • 4→34 \to 34→3
    • 4→24 \to 24→2
    • 4→14 \to 14→1
    • and later 3→23 \to 23→2, 3→13 \to 13→1, 2→12 \to 12→1

    We check which one lies in visible region.

  6. Compute wavelength for 4→34 \to 34→3 transition

    Energy emitted: ΔE=54.4(132−142)\Delta E=54.4\left(\frac{1}{3^2}-\frac{1}{4^2}\right)ΔE=54.4(321​−421​) =54.4(19−116)=54.4\left(\frac{1}{9}-\frac{1}{16}\right)=54.4(91​−161​) =54.4(16−9144)=54.4⋅7144=54.4\left(\frac{16-9}{144}\right)=54.4\cdot\frac{7}{144}=54.4(14416−9​)=54.4⋅1447​ ΔE≈2.64 eV\Delta E\approx 2.64\text{ eV}ΔE≈2.64 eV

    Then λ=1240 eV nm2.64 eV≈470 nm=4.7×10−7 m\lambda=\frac{1240\text{ eV nm}}{2.64\text{ eV}}\approx 470\text{ nm}=4.7\times 10^{-7}\text{ m}λ=2.64 eV1240 eV nm​≈470 nm=4.7×10−7 m

    This is in the visible region.

  7. Check other transitions briefly

    • 4→24\to24→2: ΔE=54.4(14−116)=10.2 eV\Delta E=54.4\left(\frac14-\frac1{16}\right)=10.2\text{ eV}ΔE=54.4(41​−161​)=10.2 eV λ≈124010.2≈122 nm\lambda\approx\frac{1240}{10.2}\approx 122\text{ nm}λ≈10.21240​≈122 nm UV

    • 4→14\to14→1 and 3→13\to13→1, 2→12\to12→1 are even higher energies, hence UV.

    • 3→23\to23→2: ΔE=54.4(14−19)=54.4⋅536≈7.56 eV\Delta E=54.4\left(\frac14-\frac19\right)=54.4\cdot\frac{5}{36}\approx 7.56\text{ eV}ΔE=54.4(41​−91​)=54.4⋅365​≈7.56 eV λ≈164 nm\lambda\approx 164\text{ nm}λ≈164 nm UV

    Therefore, the only visible emission is due to 4→34\to34→3

  8. Match with options

    λ≈4.7×10−7 m\lambda\approx 4.7\times 10^{-7}\text{ m}λ≈4.7×10−7 m Closest option is 4.8×10−7 m4.8\times10^{-7}\text{ m}4.8×10−7 m

    So the correct option is C.

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