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Question

(B) The energy of a photon emitted during a transition in a hydrogen atom is given by the formula: $E = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2}

Vedclass · Accepted Answer

$3.4$

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(B) The energy of a photon emitted during a transition in a hydrogen atom is given by the formula: $E = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] eV$.For the Balmer series,the transition ends at the energy level $n_1 = 2$.The highest energy photon corresponds to the transition from the highest possible energy level,which is $n_2 = \infty$.Substituting these values into the formula:$E = 13.6 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] eV$$E = 13.6 \left[ \frac{1}{4} - 0 \right] eV$$E = \frac{13.6}{4} eV = 3.4 eV$.Therefore,the energy of the highest energy photon of the Balmer series is $3.4 eV$.

The energy of the highest energy photon of the Balmer series of the hydrogen spectrum is close to $... eV$.

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