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Atoms and Nuclei question

2024 · 4 Apr · Shift 1 · Q83
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Atoms and Nuclei question

2024 · 4 Apr · Shift 1 · Q83

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
A hydrogen atom changes its state from n=3n=3n=3 to n=2n=2n=2. Due to recoil, the percentage change in the wave length of emitted light is approximately 1×10−n1 \times 10^{-n}1×10−n. The value of nnn is ‾\underline{\hspace{2cm}}​. [Given Rhc =13.6 eV,hc=1242 eV nm,h=6.6×10−34 J s=13.6 \mathrm{~eV}, \mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}=13.6 eV,hc=1242 eV nm,h=6.6×10−34 J s mass of the hydrogenatom =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg}=1.6×10−27 kg]
Numerical answer
View written solutionFree

Correct answer: 7

  1. Energy of emitted photon without recoil

For the transition n=3→n=2n=3 \to n=2n=3→n=2 in hydrogen,

ΔE=13.6(122−132) eV=13.6(14−19)\Delta E = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \text{ eV} =13.6\left(\frac{1}{4}-\frac{1}{9}\right)ΔE=13.6(221​−321​) eV=13.6(41​−91​) 14−19=536\frac{1}{4}-\frac{1}{9}=\frac{5}{36}41​−91​=365​

So,

ΔE=13.6×536≈1.89 eV\Delta E=13.6\times \frac{5}{36}\approx 1.89\text{ eV}ΔE=13.6×365​≈1.89 eV
  1. Wavelength of emitted light ignoring recoil

Using hc=1242 eV nmhc=1242\,\text{eV nm}hc=1242eV nm,

λ=hcΔE=12421.89 nm≈657 nm\lambda = \frac{hc}{\Delta E} = \frac{1242}{1.89} \text{ nm} \approx 657 \text{ nm}λ=ΔEhc​=1.891242​ nm≈657 nm
  1. Recoil energy of hydrogen atom

When the atom emits a photon, it recoils with momentum equal to photon momentum:

p=hλp=\frac{h}{\lambda}p=λh​

Hence recoil energy,

Er=p22M=12M(hλ)2E_r=\frac{p^2}{2M}=\frac{1}{2M}\left(\frac{h}{\lambda}\right)^2Er​=2Mp2​=2M1​(λh​)2

Substitute values:

  • h=6.6×10−34 J sh=6.6\times 10^{-34}\,\text{J s}h=6.6×10−34J s
  • λ=657×10−9 m\lambda=657\times 10^{-9}\,\text{m}λ=657×10−9m
  • M=1.6×10−27 kgM=1.6\times 10^{-27}\,\text{kg}M=1.6×10−27kg

First,

p=6.6×10−34657×10−9≈1.0×10−27 kg m/sp=\frac{6.6\times 10^{-34}}{657\times 10^{-9}}\approx 1.0\times 10^{-27}\,\text{kg m/s}p=657×10−96.6×10−34​≈1.0×10−27kg m/s

Then,

Er=(1.0×10−27)22×1.6×10−27=10−543.2×10−27≈3.1×10−28 JE_r=\frac{(1.0\times 10^{-27})^2}{2\times 1.6\times 10^{-27}} =\frac{10^{-54}}{3.2\times 10^{-27}} \approx 3.1\times 10^{-28}\,\text{J}Er​=2×1.6×10−27(1.0×10−27)2​=3.2×10−2710−54​≈3.1×10−28J

Convert this into eV:

Er≈3.1×10−281.6×10−19 eV≈1.9×10−9 eVE_r \approx \frac{3.1\times 10^{-28}}{1.6\times 10^{-19}} \text{ eV} \approx 1.9\times 10^{-9}\text{ eV}Er​≈1.6×10−193.1×10−28​ eV≈1.9×10−9 eV
  1. Fractional change in wavelength

Because of recoil, the emitted photon energy is reduced by ErE_rEr​. For small change,

Δλλ≈ErΔE\frac{\Delta \lambda}{\lambda} \approx \frac{E_r}{\Delta E}λΔλ​≈ΔEEr​​

So,

Δλλ≈1.9×10−91.89≈10−9\frac{\Delta \lambda}{\lambda} \approx \frac{1.9\times 10^{-9}}{1.89} \approx 10^{-9}λΔλ​≈1.891.9×10−9​≈10−9

Thus percentage change is

Δλλ×100≈10−9×100=10−7\frac{\Delta \lambda}{\lambda}\times 100 \approx 10^{-9}\times 100=10^{-7}λΔλ​×100≈10−9×100=10−7

So the percentage change is approximately

1×10−71\times 10^{-7}1×10−7

Hence,

n=7n=7n=7
  1. Comparison with stored answer

Stored correct answer = 777.

Our derived answer also gives 777.

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