Atoms & Nuclei

JEE Main level, numerical answer, medium

45 of 239

Question 45 of 239

An atom absorbs a photon of wavelength 500nm500nm and emits another photon of wavelength 600nm600nm. The net energy absorbed by the atom in this process is n×10−4eVn\times {10}^{-4}eV. The value ofnnis [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6×10−34Jsh=6.6\times {10}^{-34}Js andc=3×108ms−1and c=3×108 m s-1 ).

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Answer: 4125

The energy of a photon is given by E=hcλE=\dfrac{hc}{\lambda }.
It is given that λ1=500nm{\lambda }_{1}=500nm, λ2=600nm{\lambda }_{2}=600nm.
The net energy absorbed is
ΔE=hcλ1−hcλ2=hc10−9(1500−1600)\Delta E=\dfrac{hc}{{\lambda }_{1}}-\dfrac{hc}{{\lambda }_{2}}=\dfrac{hc}{{10}^{-9}}(\dfrac{1}{500}-\dfrac{1}{600})
=6.6×10−34×3×108×100500×600×10−9=\dfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}\times 100}{500\times 600\times {10}^{-9}}
=6.6×330×10−19J=\dfrac{6.6\times 3}{30}\times {10}^{-19}J
1eV=1.6×10−19J1eV=1.6\times {10}^{-19}J
=6.6×330×1.6eV=\dfrac{6.6\times 3}{30\times 1.6}eV
=4125×10−4eV=4125\times {10}^{-4}eV

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