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Aloiza [94]
3 years ago
14

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

on emitted is:
Chemistry
1 answer:
viktelen [127]3 years ago
3 0

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

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

Formula used :

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

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

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

Energy change transition from n = 1 to n = 3 occurs.

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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