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Energy Levels

Energy E of a...

Question

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 n = 3 state to n = 2 state of hydrogen is approximately

1.9 eV

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1.5 eV

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0.85 eV

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3.4 eV

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Solution

The correct option is A 1.9 eV
$Δ E = 13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = 13.6 \times \frac{5}{36} = 1.9 e V$

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Similar questions

E = \frac{- 13.6}{n^{2}}

n

E_{n} = - \frac{13.6 Z^{2}}{n^{2}}

E_{m a x} = 52.224 e V

E_{m i n} = 1.224 e V

n

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

n

13.6 e V

(n = 2)

n^{t h}

E_{n} = - \frac{13.6}{n^{2}} Z^{2}

E_{m a x} = 52.224 e V

E_{m i n} = 1.224 e V

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