TS EAMCET · Physics · Atomic Physics
The frequency of light emitted, when the electron makes transition from the level of principle quantum number \(n=2\) to the level with \(n=1\) is (Take, the ionization energy of hydrogen to be \(13.6 \mathrm{eV}\) and \(h \simeq 4 \times 10^{-15} \mathrm{eV}\)-s)
- A \(2.55 \times 10^{15} \mathrm{~Hz}\)
- B \(1.7 \times 10^{15} \mathrm{~Hz}\)
- C \(3.4 \times 10^{15} \mathrm{~Hz}\)
- D \(5.1 \times 10^{15} \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(2.55 \times 10^{15} \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
As ionisation energy is \(13.6 \mathrm{eV}\), so its negative will be the energy of atom in ground state \((n=1)\) hence \(B_1=-136 \mathrm{eV}\) Energy in second orbit \((n=1)\) \(B_2=-\frac{13.6}{2^2}=-3.4 \mathrm{eV}\) When electron jumps from \(n=2\) to \(n=1\), then energy…
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