Question

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

Answer 1

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* (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* (1^2)/(1^2)eV=-13.6eV`

Energy of n = 2 in an hydrogen atom:

`E_3=-13.6* (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