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Structure of Atom question

2022 · 28 Jul · Shift 2 · Q15
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Structure of Atom question

2022 · 28 Jul · Shift 2 · Q15

JEE MainChemistryStructure of AtomNumerical+4 / −1
If the wavelength for an electron emitted from H\mathrm{H}H-atom is 3.3×10−10 m3.3 \times 10^{-10} \mathrm{~m}3.3×10−10 m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is ‾\underline{\hspace{2cm}}​ times. (Nearest integer) [\left[\right.[ Given :h=6.626×10−34 J s: \mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}:h=6.626×10−34 J s] Mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg}=9.1×10−31 kg
Numerical answer
View written solutionFree

Correct answer: 2

  1. Interpretation of the question

An electron is emitted from a hydrogen atom with wavelength

λ=3.3×10−10 m\lambda = 3.3\times 10^{-10}\,\text{m}λ=3.3×10−10m

This is the de Broglie wavelength of the emitted electron.

We need the ratio:

energy absorbed by electron in ground stateminimum energy required for escape\frac{\text{energy absorbed by electron in ground state}}{\text{minimum energy required for escape}}minimum energy required for escapeenergy absorbed by electron in ground state​

For hydrogen in ground state, the minimum energy required for escape is the ionization energy:

Eion=13.6 eVE_\text{ion} = 13.6\,\text{eV}Eion​=13.6eV

If after absorption the electron escapes with some kinetic energy KKK, then

Eabsorbed=Eion+KE_\text{absorbed} = E_\text{ion} + KEabsorbed​=Eion​+K

So required ratio is

EabsorbedEion=13.6+K13.6\frac{E_\text{absorbed}}{E_\text{ion}} = \frac{13.6 + K}{13.6}Eion​Eabsorbed​​=13.613.6+K​
  1. Find kinetic energy from de Broglie wavelength

Using

λ=hmv\lambda = \frac{h}{mv}λ=mvh​

so momentum is

p=hλp = \frac{h}{\lambda}p=λh​

Hence kinetic energy:

K=p22m=h22mλ2K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}K=2mp2​=2mλ2h2​

Substitute the values:

h=6.626×10−34 J sh = 6.626\times 10^{-34}\,\text{J s}h=6.626×10−34J s m=9.1×10−31 kgm = 9.1\times 10^{-31}\,\text{kg}m=9.1×10−31kg λ=3.3×10−10 m\lambda = 3.3\times 10^{-10}\,\text{m}λ=3.3×10−10m

Therefore,

K=(6.626×10−34)22(9.1×10−31)(3.3×10−10)2K = \frac{(6.626\times 10^{-34})^2}{2(9.1\times 10^{-31})(3.3\times 10^{-10})^2}K=2(9.1×10−31)(3.3×10−10)2(6.626×10−34)2​

Now,

(6.626×10−34)2=4.390×10−67(6.626\times 10^{-34})^2 = 4.390\times 10^{-67}(6.626×10−34)2=4.390×10−67 (3.3×10−10)2=10.89×10−20=1.089×10−19(3.3\times 10^{-10})^2 = 10.89\times 10^{-20} = 1.089\times 10^{-19}(3.3×10−10)2=10.89×10−20=1.089×10−19 2mλ2=2(9.1×10−31)(1.089×10−19)2m\lambda^2 = 2(9.1\times 10^{-31})(1.089\times 10^{-19})2mλ2=2(9.1×10−31)(1.089×10−19) =1.98198×10−49= 1.98198\times 10^{-49}=1.98198×10−49

Thus,

K=4.390×10−671.98198×10−49approx2.215×10−18 JK = \frac{4.390\times 10^{-67}}{1.98198\times 10^{-49}} approx 2.215\times 10^{-18}\,\text{J}K=1.98198×10−494.390×10−67​approx2.215×10−18J

Convert into eV using

1 eV=1.6×10−19 J1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}1eV=1.6×10−19J K=2.215×10−181.6×10−19≈13.84 eVK = \frac{2.215\times 10^{-18}}{1.6\times 10^{-19}} \approx 13.84\,\text{eV}K=1.6×10−192.215×10−18​≈13.84eV
  1. Calculate absorbed energy

Since electron escapes from ground state,

Eabsorbed=13.6+13.84=27.44 eVE_\text{absorbed} = 13.6 + 13.84 = 27.44\,\text{eV}Eabsorbed​=13.6+13.84=27.44eV
  1. Required ratio
Ratio=27.4413.6≈2.02\text{Ratio} = \frac{27.44}{13.6} \approx 2.02Ratio=13.627.44​≈2.02

Nearest integer:

2\boxed{2}2​
  1. Comparison with stored answer

Stored correct answer = 222

Our derived answer = 222

So the stored answer is correct.

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