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

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell

i think it is true since you can test the answer and can use observations to complete it.