JEE Mains · Physics · STD 12 - 12. atoms
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 \(\left.c =3 \times 10^8\,m / s \right)\)
- A \(4124\)
- B \(4125\)
- C \(4123\)
- D \(4122\)
Answer & Solution
Correct Answer
(B) \(4125\)
Step-by-step Solution
Detailed explanation
\(E = E _1- E _2=\frac{ hc }{\lambda_1}-\frac{ hc }{\lambda_2}= hc \left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)\) \(=6.6 \times 10^{-34} \times 3 \times 10^8\left(\frac{1}{500 \times 10^{-9}}-\frac{1}{600 \times 10^{-9}}\right)\) \(=6.6 \times 10^{-20}\,J\)…
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