A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system. The enthalpies of these reactions are less than zero, and are therefore exothermic reactions
The answer would be 6 e-. This is becuase you are turning a charge of -4 into a +2. In order to do this, you transfer 4 electrons for a neutral charge, and an additional 2 for a charge of +2.
This makes a total charge of +2, and the total transferred electrons 6 e-
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