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Dual Nature of Radiation question

2023 · 13 Apr · Shift 2 · Q63
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  5. /2023 · 13 Apr · Shift 2 · Q63

Dual Nature of Radiation question

2023 · 13 Apr · Shift 2 · Q63

JEE MainPhysicsDual Nature of RadiationNumerical+4 / −1
An atom absorbs a photon of wavelength 500 nm500 \mathrm{~nm}500 nm and emits another photon of wavelength 600 nm600 \mathrm{~nm}600 nm. The net energy absorbed by the atom in this process is n×10−4 eVn \times 10^{-4} ~\mathrm{eV}n×10−4 eV. The value of n is ‾\underline{\hspace{2cm}}​. [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6×10−34 Js\mathrm{h}=6.6 \times 10^{-34} ~\mathrm{Js}h=6.6×10−34 Js and c=3×108 m/s\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}c=3×108 m/s )
Numerical answer
View written solutionFree

Correct answer: 4125

  1. Energy of a photon

The energy of a photon is

E=hcλE=\frac{hc}{\lambda}E=λhc​

Here,

  • h=6.6×10−34 J sh=6.6\times 10^{-34}\,\text{J s}h=6.6×10−34J s
  • c=3×108 m/sc=3\times 10^8\,\text{m/s}c=3×108m/s

So,

hc=(6.6×10−34)(3×108)=19.8×10−26=1.98×10−25 J mhc=(6.6\times 10^{-34})(3\times 10^8)=19.8\times 10^{-26}=1.98\times 10^{-25}\,\text{J m}hc=(6.6×10−34)(3×108)=19.8×10−26=1.98×10−25J m

  1. Energy absorbed by atom

The atom absorbs a photon of wavelength 500 nm500\,\text{nm}500nm.

λ1=500×10−9 m=5×10−7 m\lambda_1=500\times 10^{-9}\,\text{m}=5\times 10^{-7}\,\text{m}λ1​=500×10−9m=5×10−7m

Hence,

E1=1.98×10−255×10−7=0.396×10−18=3.96×10−19 JE_1=\frac{1.98\times 10^{-25}}{5\times 10^{-7}}=0.396\times 10^{-18}=3.96\times 10^{-19}\,\text{J}E1​=5×10−71.98×10−25​=0.396×10−18=3.96×10−19J

  1. Energy emitted by atom

The atom emits a photon of wavelength 600 nm600\,\text{nm}600nm.

λ2=600×10−9 m=6×10−7 m\lambda_2=600\times 10^{-9}\,\text{m}=6\times 10^{-7}\,\text{m}λ2​=600×10−9m=6×10−7m

Thus,

E2=1.98×10−256×10−7=0.33×10−18=3.3×10−19 JE_2=\frac{1.98\times 10^{-25}}{6\times 10^{-7}}=0.33\times 10^{-18}=3.3\times 10^{-19}\,\text{J}E2​=6×10−71.98×10−25​=0.33×10−18=3.3×10−19J

  1. Net energy absorbed

Since the atom first absorbs and then emits, net energy gained by atom is

ΔE=E1−E2\Delta E=E_1-E_2ΔE=E1​−E2​

ΔE=(3.96−3.3)×10−19=0.66×10−19=6.6×10−20 J\Delta E=(3.96-3.3)\times 10^{-19}=0.66\times 10^{-19}=6.6\times 10^{-20}\,\text{J}ΔE=(3.96−3.3)×10−19=0.66×10−19=6.6×10−20J

  1. 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

we get

ΔE=6.6×10−201.6×10−19 eV\Delta E=\frac{6.6\times 10^{-20}}{1.6\times 10^{-19}}\,\text{eV}ΔE=1.6×10−196.6×10−20​eV

ΔE=4.125×10−1 eV=0.4125 eV\Delta E=4.125\times 10^{-1}\,\text{eV}=0.4125\,\text{eV}ΔE=4.125×10−1eV=0.4125eV

Now the question says

ΔE=n×10−4 eV\Delta E=n\times 10^{-4}\,\text{eV}ΔE=n×10−4eV

So,

0.4125=n×10−40.4125=n\times 10^{-4}0.4125=n×10−4

n=0.4125×104=4125n=0.4125\times 10^4=4125n=0.4125×104=4125

  1. Final answer

n=4125\boxed{n=4125}n=4125​

The derived answer matches the stored correct answer.

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