Answer
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Work Step by Step
The 12.3 V potential difference means the electrons have a kinetic energy of 12.3 eV. When they collide with the hydrogen atom, they may transfer this energy. Starting from the ground state, the H atom might have as much as $-13.6eV+12.3eV=-1.3eV$ of energy. The energy level diagram in Figure 27–29 shows us that this corresponds to exciting the H atom up to the n = 3 state.
As the atom returns to the ground state, there are 3 possible transitions.
Equation 27-15b says that for hydrogen, the energy of a level is $E_n=\frac{-13.6eV}{n^2}$. Combine this with equation 27–10. $E_1$, $E_2$, and $E_3$ are shown in the textbook. Calculate the wavelengths.
From n = 3 to n = 2.
$$\lambda=\frac{hc}{E_3-E_2}=\frac{1240eV\cdot nm}{-1.51eV-(-3.40eV)}=650nm$$
From n = 3 to n = 1.
$$\lambda=\frac{hc}{E_4-E_2}=\frac{1240eV\cdot nm}{-1.51eV-(-13.6eV)}=102nm$$
From n = 2 to n = 1.
$$\lambda=\frac{hc}{E_2-E_1}=\frac{1240eV\cdot nm}{-3.40eV-(-13.6eV)}=122nm$$