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Atoms and Nuclei question

2009 · Shift 0 · Q52
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Atoms and Nuclei question

2009 · Shift 0 · Q52

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The transition from the state n=4n=4n=4 to n=3n=3n=3 in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from :
  1. A
    3→23 \to 23→2
  2. B
    4→24 \to 24→2
  3. C
    5→45 \to 45→4
  4. D
    2→12 \to 12→1
View written solutionFree

Correct answer: C

  1. For a hydrogen-like atom, the emitted photon energy in a transition ni→nfn_i \to n_fni​→nf​ is
ΔE=13.6Z2(1nf2−1ni2) eV,ni>nf.\Delta E = 13.6 Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \text{ eV}, \qquad n_i>n_f.ΔE=13.6Z2(nf2​1​−ni2​1​) eV,ni​>nf​.

Since wavelength is related by

λ=hcΔE,\lambda = \frac{hc}{\Delta E},λ=ΔEhc​,

a smaller energy gap gives a larger wavelength.

  1. It is given that the transition 4→34 \to 34→3 produces ultraviolet radiation. Therefore:
  • any transition with larger energy gap than 4→34 \to 34→3 will have shorter wavelength,
  • any transition with smaller energy gap than 4→34 \to 34→3 will have longer wavelength.

Infrared has longer wavelength than ultraviolet, so we need a transition with energy gap smaller than that of 4→34 \to 34→3.

  1. Compare the quantity
(1nf2−1ni2)\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)(nf2​1​−ni2​1​)

for each option.

For the given transition 4→34 \to 34→3:

Δ=132−142=19−116=7144.\Delta = \frac{1}{3^2}-\frac{1}{4^2} = \frac{1}{9}-\frac{1}{16} = \frac{7}{144}.Δ=321​−421​=91​−161​=1447​.
  1. Now evaluate each option:
  • A: 3→23 \to 23→2
122−132=14−19=536.\frac{1}{2^2}-\frac{1}{3^2} = \frac{1}{4}-\frac{1}{9} = \frac{5}{36}.221​−321​=41​−91​=365​.

Since

536>7144,\frac{5}{36} > \frac{7}{144},365​>1447​,

this has larger energy gap, hence shorter wavelength than 4→34 \to 34→3. Not infrared.

  • B: 4→24 \to 24→2
122−142=14−116=316.\frac{1}{2^2}-\frac{1}{4^2} = \frac{1}{4}-\frac{1}{16} = \frac{3}{16}.221​−421​=41​−161​=163​.

Since

316>7144,\frac{3}{16} > \frac{7}{144},163​>1447​,

this also has larger energy gap. Not infrared.

  • C: 5→45 \to 45→4
142−152=116−125=9400.\frac{1}{4^2}-\frac{1}{5^2} = \frac{1}{16}-\frac{1}{25} = \frac{9}{400}.421​−521​=161​−251​=4009​.

Now

9400<7144,\frac{9}{400} < \frac{7}{144},4009​<1447​,

so this has smaller energy gap, hence longer wavelength than 4→34 \to 34→3. Therefore it will lie in the infrared region.

  • D: 2→12 \to 12→1
1−14=34,1-\frac{1}{4} = \frac{3}{4},1−41​=43​,

which is much larger than 7144\frac{7}{144}1447​. So this has much shorter wavelength. Not infrared.

  1. Hence the only transition that gives radiation of longer wavelength than the given ultraviolet transition is
5→4.5 \to 4.5→4.

Therefore, the correct option is C.

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