If electron falls from n = 3 to n = 2, then emitted energy is
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If electron falls from n = 3 to n = 2, then emitted energy is
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Answer:
ΔE=E3−E2=13.6[1(2)2−1(3)2]=1.9eV
Answer:
The energy emitted when an electron transitions from a higher energy level (n = 3) to a lower energy level (n = 2) can be calculated using the formula for the energy difference between energy levels in an atom.
The formula for the energy of a photon emitted or absorbed during an electron transition between two energy levels in an atom is given by:
\[ \Delta E = E_2 - E_3 = -R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
Where:
- \(\Delta E\) is the energy difference between the initial (n = 3) and final (n = 2) energy levels.
- \(E_2\) is the energy of the final state (n = 2).
- \(E_3\) is the energy of the initial state (n = 3).
- \(R_H\) is the Rydberg constant (\(2.18 \times 10^{-18}\) J)).
- \(n_1\) and \(n_2\) are the quantum numbers representing the energy levels.
Substituting the values into the formula:
\(\Delta E = -R_H \left( \frac{1}{{2^2}} - \frac{1}{{3^2}} \right)\)
\(\Delta E = -R_H \left( \frac{1}{4} - \frac{1}{9} \right)\)
\(\Delta E = -R_H \left( \frac{9 - 4}{36} \right)\)
\(\Delta E = -R_H \times \frac{5}{36}\)
Using the value of the Rydberg constant \(R_H \approx 2.18 \times 10^{-18}\) J, you can calculate the numerical value of \(\Delta E\).
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