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Atoms and Nuclei question

2021 · 26 Aug · Shift 1 · Q52
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  5. /2021 · 26 Aug · Shift 1 · Q52

Atoms and Nuclei question

2021 · 26 Aug · Shift 1 · Q52

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
A particular hydrogen like ion emits radiation of frequency 2.92 ×\times× 1015 Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be :
  1. A
    0.44 ×\times× 1015
  2. B
    6.57 ×\times× 1015
  3. C
    4.38 ×\times× 1015
  4. D
    2.46 ×\times× 1015
View written solutionFree

Correct answer: D

  1. Use the Bohr frequency relation for a hydrogen-like ion

For a transition from nin_ini​ to nfn_fnf​ in a hydrogen-like ion,

ν=RcZ2(1nf2−1ni2)\nu = RcZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)ν=RcZ2(nf2​1​−ni2​1​)

where RRR is the Rydberg constant, ccc is speed of light, and ZZZ is atomic number.

So frequency is proportional to

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

for the same ion.

  1. Given transition: n=3n=3n=3 to n=1n=1n=1
ν31=2.92×1015 Hz\nu_{31} = 2.92 \times 10^{15}\,\text{Hz}ν31​=2.92×1015Hz

Also,

ν31∝(1−19)=89\nu_{31} \propto \left(1-\frac{1}{9}\right)=\frac{8}{9}ν31​∝(1−91​)=98​
  1. Required transition: n=2n=2n=2 to n=1n=1n=1
ν21∝(1−14)=34\nu_{21} \propto \left(1-\frac{1}{4}\right)=\frac{3}{4}ν21​∝(1−41​)=43​
  1. Take ratio
ν21ν31=3489=34⋅98=2732\frac{\nu_{21}}{\nu_{31}}=\frac{\frac{3}{4}}{\frac{8}{9}}=\frac{3}{4}\cdot\frac{9}{8}=\frac{27}{32}ν31​ν21​​=98​43​​=43​⋅89​=3227​

Hence,

ν21=ν31⋅2732\nu_{21}=\nu_{31}\cdot \frac{27}{32}ν21​=ν31​⋅3227​ ν21=2.92×1015×2732\nu_{21}=2.92\times 10^{15}\times \frac{27}{32}ν21​=2.92×1015×3227​ ν21=2.46375×1015 Hz\nu_{21}=2.46375\times 10^{15}\,\text{Hz}ν21​=2.46375×1015Hz ν21≈2.46×1015 Hz\nu_{21}\approx 2.46\times 10^{15}\,\text{Hz}ν21​≈2.46×1015Hz
  1. Match with options

This corresponds to:

Option D: 2.46×10152.46 \times 10^{15}2.46×1015 Hz

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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