Chapter 38 - Quantization - Exercises and Problems - Page 1152: 22

(a) $\lambda_{31} = 310~ nm$ $\lambda_{32} = 500~ nm$ $\lambda_{21} = 830~ nm$ (b) $\lambda_{13} = 310~ nm$ $\lambda_{12} = 830~ nm$

(a) One wavelength is associated with the jump from $n = 3$ to $n = 1$. In this case, the energy difference is: $~~\Delta E = 4.0~eV-0.0~eV = 4.0~eV$ We can find the wavelength: $E = \frac{h~c}{\lambda}$ $\lambda_{31} = \frac{h~c}{E}$ $\lambda_{31} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(4.0~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda _{31}= 3.1\times 10^{-7}~m$ $\lambda_{31} = 310~ nm$ One wavelength is associated with the jump from $n = 3$ to $n = 2$. In this case, the energy difference is: $~~\Delta E = 4.0~eV-1.5~eV = 2.5~eV$ We can find the wavelength: $E = \frac{h~c}{\lambda}$ $\lambda_{32} = \frac{h~c}{E}$ $\lambda_{32} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.5~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda_{32} = 5.0\times 10^{-7}~m$ $\lambda_{32} = 500~ nm$ One wavelength is associated with the jump from $n = 2$ to $n = 1$. In this case, the energy difference is: $~~\Delta E = 1.5~eV-0.0~eV = 1.5~eV$ We can find the wavelength: $E = \frac{h~c}{\lambda}$ $\lambda_{21} = \frac{h~c}{E}$ $\lambda_{21} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.5~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda_{21} = 8.3\times 10^{-7}~m$ $\lambda_{21} = 830~ nm$ (b) We can find the wavelength $\lambda_{13}$: $E = \frac{h~c}{\lambda}$ $\lambda_{13} = \frac{h~c}{E}$ $\lambda_{13} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(4.0~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda _{13}= 3.1\times 10^{-7}~m$ $\lambda_{13} = 310~ nm$ We can find the wavelength $\lambda_{12}$: $E = \frac{h~c}{\lambda}$ $\lambda_{12} = \frac{h~c}{E}$ $\lambda_{12} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.5~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda_{12} = 8.3\times 10^{-7}~m$ $\lambda_{12} = 830~ nm$