The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :
- 1.5 eV
- 13.6 eV
- 1.9 eV
- 12.1 eV
The Correct Option is D
Approach Solution - 1
For the hydrogen atom, the transition from \(n = 1\) to \(n = 3\) gives the radiation in the Balmer series:
\[ \Delta E = 12.1 \, \text{eV} \]
Thus, the correct answer is Option (4).
Approach Solution -2
A hydrogen atom in the ground state (n=1) must first absorb energy to reach an excited state from which a transition to n=2 (first excited state) results in the emission of a Balmer series spectral line. We need to find the minimum energy required for this process.
Concept Used:
The energy of an electron in the nth orbit of a hydrogen atom is given by the Bohr model formula:
\[ E_n = -\frac{13.6}{n^2} \ \text{eV} \]
Step-by-Step Solution:
Step 1: Write the energies of the relevant levels.
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \ \text{eV} \] \[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \ \text{eV} \]
Step 2: Calculate the energy difference for the transition from n=1 to n=3.
\[ \Delta E = E_3 - E_1 = \left(-\frac{13.6}{9}\right) - (-13.6) = 13.6 - \frac{13.6}{9} \] \[ \Delta E = 13.6 \left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} \]
Step 3: Compute the numerical value.
\[ \Delta E = \frac{108.8}{9} \approx 12.0889 \ \text{eV} \]
This value is approximately 12.1 eV.
Thus, the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is nearly 12.1 eV.
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