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Chemistry · Structure of atom
हिन्दीCalculate the frequency of the radiation emitted when an electron jumps from the \( n = 5 \) state to the \( n = 2 \) state in a hydrogen atom (\( R_H = 2.18 \times 10^{-18} \text{ J} \), \( h = 6.626 \times 10^{-34} \text{ J s} \)).
Correct answer · Option 4
\( 6.91 \times 10^{14} \text{ Hz} \)
Explanation
Using \( \nu = \frac{\Delta E}{h} = \frac{R_H}{h} (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \), substituting \( n_1=2 \) and \( n_2=5 \) yields \( 6.91 \times 10^{14} \text{ Hz} \).
