The ionization energy of a hydrogen-like atom is 14.4 eV. The amount of energy released whenelectron jumps from the fourth orbit to the first orbit in this atom is (in eV)

The ionization energy of a hydrogen-like atom is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞). Given that the ionization energy is 14.4 eV, we can use the formula for the energy levels of a hydrogen-like atom:

E_n = -13.6 Z^2 / n^2 eV

where Z is the atomic number and n is the principal quantum number. For hydrogen-like atoms, Z=1. So, the energy of the ground state (n=1) is:

E_1 = -13.6 * 1^2 / 1^2 = -13.6 eV

Given that the ionization energy is 14.4 eV, the energy of the atom in its ionized state (n=∞) is:

E_∞ = E_1 + Ionization energy = -13.6 eV + 14.4 eV = 0.8 eV

Now, we can calculate the energy of the fourth state (n=4):

E_4 = -13.6 * 1^2 / 4^2 = -13.6 / 16 = -0.85 eV

The energy released when the electron jumps from the fourth orbit to the first orbit is the difference in energy between these two states:

ΔE = E_1 - E_4 = -13.6 eV - (-0.85 eV) = -12.75 eV

The negative sign indicates that energy is released. Therefore, the amount of energy released is 12.75 eV.