QNEET-2023-Ph-341: NEET 2023 Physics - Class 12 NEET-2023-Ph-341 Solution: The ground state energy of hydrogen atom is -13.6 eV. The…

The energy levels of a hydrogen atom are given by the formula : E n = _ 13.6   eV n 2 where E n $E_n$ E n _ is the energy of the n $n$ n -th energy level. The ground state of hydrogen ( n = 1 $n=1$ n = 1 ) has an energy of _ 13.6   e V $-13.6 ~\mathrm{eV}$ _ 13.6   eV as mentioned, which means that it would take + 13.6   e V $+13.6 ~\mathrm{eV}$ + 13.6   eV to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy. The second excited state of hydrogen is when n = 3 $n=3$ n = 3 (as n = 1 $n=1$ n = 1 is the ground state and n = 2 $n=2$ n = 2 is the first excited state). Thus, the energy of the second excited state is : E 3 = _ 13.6   eV 3 2 = _ 13.6   eV 9 = _ 1.51   eV Since ionization implies moving the electron from its current energy level to 0 $0$ 0 energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take + 1.51   e V $+1.51 ~\mathrm{eV}$ + 1.51   eV to ionize a hydrogen atom from its second excited state. So, the correct answer is Option C : 1.51   e V $1.51 ~\mathrm{eV}$ 1.51   eV .