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

Explanation :

First we have to calculate the orbit of hydrogen atom.

Formula used :

where,

= energy of orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

Energy of n = 2 in an hydrogen atom:

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

Let energy change be E.

The negative sign indicates that energy of the photon emitted.

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