Problems

#05

This is a unit conversion problem. It can be approached in a couple different ways, but the key is to compare the two photons in the same units. I choose to convert the radio frequency to an energy to make the comparison. The frequency is $\nu=100$ MHz, or $1\times 10^8$ Hz. The conversion to energy makes use of the equation $E=h\nu = (6.63\times 10^{-34}) \times (1\times 10^8)= 6.63\times 10^{-26}$ Joules (MKS units). An electron volt (eV) is $1.60\times 10^{-19}$ Joules. That means that the radio photon has an energy of only $(6.63\times 10^{-26}) /1.60\times 10^{-19} = 4.14\times 10^{-7}$ eV. This means that the energy of the X-ray photon is $100/4.14\times 10^{-7} = 2.41\times 10^8$ times bigger than the radio photon. But frequency is related to energy; in fact, $\nu \propto E$, so the X-ray photon's frequency is likewise that many times bigger than the radio photon. Wavelength $\lambda$, on the other hand, is inversely related to frequency and to energy. This means as $E$ or $\nu$ increase by some factor, $\lambda$ decreases by the same factor (and vice versa). Radio wavelengths are much bigger than X-ray wavelengths. In this example the radio photon is $2.41\times 10^8$ times bigger in wavelength than the X-ray photon.

#07

Use ``More Precisely 4-1'' in chapter 4. There are a couple of steps. The energy from level 1 to 10 is $E_{10} = 13.6 (1-1/100) = 13.464$ eV. The energy from level 1 to 9 is $E_9 = 13.6 (1-1/81) = 13.432$ eV. (Note that I am keeping extra significant figures because I have to substract two numbers that are nearly equal.). So the energy from 9 to 10 must be $E = E_{10}-E_9 = 0.033$ eV. Convert to MKS units, giving $0.033 \times (1.60\times 10^{-19}= 5.3 \times 10^{-21}$ J.

Now we use $E=h\nu$ to get frequency. Rearranging, $\nu = E/h = 5.3 \times 10^{-21}/6.63\times 10^{-34} = 8.0\times 10^{12}$ Hz.

To get wavelength we use $\lambda \times \nu = c$. Rearranging, $\lambda = c/\nu = 3\times 10^8/8.0\times 10^{12}=3.8\times 10^{-5}$ m. I like Angstroms, so this becomes 380,000 Å, which is also 38 microns, far in the infrared.

For the next parts, the trick is to realize that $E = E_n - E_m$, where $n$ and $m$ are the integer values of the levels being used. The answer in symbols is $E=13.6 (1/n^2-1/m^2)$ eV. For levels 99 to 100, the answer is $E=2.76\times 10^{-5}$ eV, and for levels 999 to 1000, the answer is $E=2.72\times 10^{-8}$ eV. With these values in eV, one repeats the steps above to find $6.66\times 10^9$ Hz and 0.0450 m in the first case, and then $6.56\times 10^6$ Hz and 45.7 m. The first wavelength is the mm band, and the second is far in the radio.