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Chapter 34: Problem 63

How much energy does it take to ionize a hydrogen atom in its first excited state?

Short Answer

Expert verified
The energy required to ionize a hydrogen atom in its first excited state is 3.4 eV.

Step by step solution

01

Understanding Ionization

Energy levels of a hydrogen atom are given by the formula: \[ E_n = -\frac{13.6 \, eV}{n^2} \], where \( n \) is the principal quantum number of the electron's state. Ionization refers to moving an electron from a state to infinity. In this case, the electron is in the first excited state, so \( n = 2 \). The energy at infinity is 0.
02

Calculate Starting and Ending Energies

The starting energy (\( E_{start} \)) and ending energy (\( E_{end} \)) are calculated. \( E_{start} \) for \( n = 2 \) is \[ E_{start} = -\frac{13.6 \, eV}{2^2} = -3.4 \, eV \] and \( E_{end} = 0 \, eV \).
03

Calculate Energy Change

The energy change required for ionization is the ending energy minus the starting energy, hence \[ \Delta E = E_{end} - E_{start} = 0 \, eV - (-3.4 \, eV) = 3.4 \, eV\]. This is the energy required to ionize the hydrogen atom from its first excited state.

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