JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2023
- Medium
- Numerical
An atom absorbs a photon of wavelength \(500\,nm\) and emits another photon of wavelength \(600\,nm\). The net energy absorbed by the atom in this process is \(n \times 10^{-4}\,eV\). The value of \(n\) is ............ [Assume the atom to be stationary during the absorption and emission process] \(\left(\right.\)Take \(h =6.6 \times 10^{-34}\,Js\) and \(\left.c =3 \times 10^8\,m / s \right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4125\)
Step-by-step Solution
Detailed explanation
\(E = E _1- E _2=\frac{ hc }{\lambda_1}-\frac{ hc }{\lambda_2}= hc \left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)\)
\(=6.6 \times 10^{-34} \times 3 \times 10^8\left(\frac{1}{500 \times 10^{-9}}-\frac{1}{600 \times 10^{-9}}\right)\)
\(=6.6 \times 10^{-20}\,J\)
\(=\frac{6.6 \times 10^{-20}}{1.6 \times 10^{-19}} eV =4.125 \times 10^{-1}\,eV\)
\(=4125 \times 10^{-4}\,eV\)
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