Energy E of a hydrogen atom with principal quantum number n is given by E=-13.6n2 eV. The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately:

E=-13.6n2eVThe energy of emitted photon = E3-E2=-13.632+13.622=13.6-19+14=13.6-4+936=13.6×536=1.9 eV.

Energy E of a hydrogen atom with principal quantum number n is given by E=-13.6n2 eV. The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately:

E=-13.6n2eVThe energy of emitted photon = E3-E2=-13.632+13.622=13.6-19+14=13.6-4+936=13.6×536=1.9 eV.

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Energy $E$ of a hydrogen atom with principal quantum number n is given by $E = \frac{- 13.6}{n^{2}} e V$ . The energy of a photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately:

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$1.5 eV$

$0.85 eV$

$3.4 eV$

$1.9 eV$

answer is D.

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$E = - \frac{13.6}{n^{2}} e V$

The energy of emitted photon $= E_{3} - E_{2} = \frac{- 13.6}{{(3)}^{2}} + \frac{13.6}{{(2)}^{2}} = 13.6 [\frac{- 1}{9} + \frac{1}{4}]$

$= 13.6 [\frac{- 4 + 9}{36}] = 13.6 \times \frac{5}{36} = 1.9 eV .$

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Energy E of a hydrogen atom with principal quantum number n is given by E=-13.6n2 eV. The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately:

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Energy $E$ of a hydrogen atom with principal quantum number n is given by $E = \frac{- 13.6}{n^{2}} e V$ . The energy of a photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately: