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Oxana [17]
3 years ago
11

Please help me, please i really need it :)

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
4 0

GGCCATAGGTCCCTTTAGCG

I believe this is correct (I used the complementary base)

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When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot
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Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

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E_n=-13.6\times \frac{Z^2}{n^2}ev

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E_n = energy of n^{th} orbit

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E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

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Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

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