In general the relationship between the frequency and energy of a photon of light is \begin{align} E = hf \end{align} where \(h\) is Plank's constant. We also know the relationship between the frequency and wavelength of light is \begin{align} c = \lambda f \end{align} where \(c\) is the speed of light. Combining these two equations and inputting our wavelength of light gives us the energy of single photon of Red light emitted by the laser as \begin{align} E&=\frac{hc}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(633\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{6.33\times10^{-7}}\text{ J}\notag \\ &=3.14\times10^{-19}\text{ J} =1.96 \text{ eV} \notag \end{align} We also note that we could also have used the fact that \(hc = 1240\text{ eV nm}\).
We apply the same calculation with the new wavelength \begin{align} E&=\frac{hc}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(590\times10^{-9})}\text{ J}\\ &= \frac{1.99\times10^{-25}}{5.90\times10^{-7}}\text{ J}\\ &=3.37\times10^{-19}\text{ J} = 2.10\text{ eV} \end{align}
The power of laser \(P\), is the energy per photon \(E\) multiplied by the number of photons emitted per unit time, \(\Phi\): \begin{equation} P = E\Phi. \end{equation} We compute \(E\) for our given wavelength \begin{align} E&=\frac{hc}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(633\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{6.33\times10^{-7}}\text{ J}\notag\\ &=3.14\times10^{-19}\text{ J}\notag \end{align} and solve for \(\Phi\) \begin{align} \Phi&=\frac{P}{E}\\ &=\frac{.001}{3.14\times10^{-19}}\text{ s}^{-1}\notag\\ &=3.18\times10^{16} \text{ s}^{-1}\notag \end{align} so we conclude that \(3.18\times10^{16}\) photons are emitted from the laser per second.
We use the same equations as the previous problem, first solving for the energy per photon \begin{align} E&=\frac{ch}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(514\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{5.14\times10^{-7}}\text{ J}\notag\\ &=3.87\times10^{-19}\text{ J}\notag \end{align} and then for \(R\) \begin{align} R&=\frac{P}{E}\\ &=\frac{.005}{3.87\times10^{-19}}\text{ s}^{-1}\notag\\ &=1.29\times10^{17} \text{ s}^{-1}\notag \end{align} so we conclude that \(1.29\times10^{17}\) photons are emitted from the laser per second.
To compare we again repeat our previous calculation by calculating the different energies per photon for the given wavelengths \begin{align} E_{red}&=\frac{ch}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(650\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{6.50\times10^{-7}}\text{ J}\notag\\ &=3.06\times10^{-19}\text{ J}\notag\\ \notag\\ E_{green}&=\frac{ch}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(560\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{5.60\times10^{-7}}\text{ J}\notag\\ &=3.55\times10^{-19}\text{ J}\notag \end{align} So it follows for an arbitrary amount of power \(P\) in units of watts \begin{align} R_{\text{red}}&=\frac{P}{E_{\text{red}}}\\ &=\frac{P}{3.06\times10^{-19}}\text{ s}^{-1}\notag\\ &=(3.27\times10^{18}\text{ s}^{-1})P\notag\\ \notag\\ R_{\text{green}}&=\frac{P}{E_{\text{green}}}\\ &=\frac{P}{3.87\times10^{-19}}\text{ s}^{-1}\notag\\ &=(2.58\times10^{18}\text{ s}^{-1})P\notag \end{align} so the red LED emits more photons per second given the same amount of power. We can calculate the ratio \begin{align} \frac{R_{\text{red}}}{R_{\text{green}}} &=\frac{(3.27\times10^{18}\text{ s}^{-1})P}{ (2.58\times10^{18}\text{ s}^{-1})P}\\ &=1.27\notag \end{align} so \(27\%\) more photons are emitted by the red LED.
We calculate the energy of the radio photons \begin{align} E &= hf\\ &= (6.63 \times 10^{-34})(8.99 \times 10^{7})\text{ J}\notag\\ &= 5.96 \times 10^{-26}\text{ J}\notag \end{align} we then calculate the rate as from previous problems \begin{align} R &= \frac{P}{E} \\ &=\frac{10000}{5.96 \times 10^{-26}} \text{ s}^{-1} \notag \\ & = 1.68 \times 10^{29}\text{ s}^{-1}\notag \end{align} so \(1.68 \times 10^{29}\) photons are emitted by the station every second.
The visible spectrum of wavelengths is about \(400\text{ nm}-700\text{ nm}\). The middle of this range is 550 nm, so I'll simplify my analysis by assuming all the photons have wavelength 550 nm. The energy of each photon is \begin{align} E_{a}&=\frac{hc}{\lambda_a}\\ &=\frac{(3\times10^8)(6.6\times10^{-34})}{(550\times10^{-9})}\text{ J}\\ &= \frac{2\times10^{-25}}{5.5\times10^{-7}}\text{ J}\\ &=3.6\times10^{-19}\text{ J}\\ \notag\\ \end{align}
the corresponding rates of photon emission from the 20 watt lightbulb \begin{align} R_{a}&=\frac{P}{E_{a}}\\ &=\frac{20}{3.6\times10^{-19}}\text{ s}^{-1}\\ &=5.6\times10^{19}\text{ s}^{-1}\\ \notag \\ \end{align}
The surface area of the sphere with radius \(100\text{ m}\) is \begin{align} S_1 = 4\pi 100^2 \text{ m}^2=130000\text{ m}^2 \end{align} and the area of the pupil is \begin{equation} S_2 = \pi \left(\frac{8}{2}\right)^2 \text{mm}^2=50 \text{ mm}^2 = 5\times10^{-7}\text{ m}^2 \end{equation} So the portion of the light from the bulb that enters the eye is approximately \begin{equation} \frac{S_2}{S_1} = \frac{5\times10^{-7}}{1.3\times10^5} =4\times10^{-12} \end{equation} So the number of photons that enters the eye is approximately, \begin{align} (4\times10^{-12})\times (5.6 \times 10^{19}) = 1.6\times10^8\text{ s}^{-1} \end{align}
This is easily visible to the human eye (the human eye could see a point source emitting \(1.5 \times 10^4\) photons per second).
By definition we have the equation for computing the wavelength of a deBoglie matter wave \begin{align} \lambda = \frac{h}{p} \end{align} where \(h\) is Planks constant and \(p\) is the magnitude of the momentum of the object in question. For a classical particle we have the definitions of kinetic energy \begin{align} K = \frac{1}{2} mv^2 \end{align} and momentum \begin{align} p = mv \end{align} where \(m\) is the mass of the object in motion and \(v\) is the speed. It follows we have the relationship \begin{align} p = \sqrt{2mK} \end{align} so for our electron in question we can compute the momentum \begin{align} p_e &= \sqrt{2m_eK_e}\\ &= \sqrt{\frac{2(511000\text{ eV})(25 \text{ eV})}{c^2}}\\ &=\frac{5.05\text{ keV}}{c}\\ &=\frac{8.01\times10^{-16}\text{ J}}{c}\\ &=2.67\times10^{-24}\frac{\text{kg m}}{\text{s}} \end{align} so the corresponding wavelength is \begin{align} \lambda_e &=\frac{h}{p_e}\\ &=\frac{6.63\times10^{-34}}{2.67\times10^{-24}} \text{ m}\\ &=2.48\times10^{-10}\text{ m}\\ &=.248 \text{ nm} \end{align}
We use the same equations as the previous part and compute in reverse, it follows that the momentum of the particle with the given wavvelength is \begin{align} p_e &= \frac{h}{\lambda_e}\\ &=\frac{6.63\times10^{-34}}{420\times10^{-9} }\frac{\text{J s}}{\text{m}}\\ &=1.58\times10^{-27}\ \frac{\text{kg m}}{\text{s}} \end{align} We then solve for our kinetic energy in terms of momentum with result \begin{align} K = \frac{p^2}{2m} \end{align} so therefore \begin{align} K_e&=\frac{p_e^2}{2m_e}\\ &=\frac{(1.58\times10^{-27})^2}{2(9.11\times10^{-31})}\text{ J}\\ &=1.37\times10^{-24}\text{ J}\\ &=8.55 \ \mu\text{eV} \end{align} a very small amount of energy.
We compute the momentum of a thermal nuetron \begin{align} p_n &= \sqrt{2m_nK_n}\\ &=\sqrt{\frac{2(939 \text{ MeV})(.04\text{ eV})}{c^2}}\\ &=\frac{\sqrt{2(939\times10^6)(.04)} \text{ eV}}{c}\\ &=\frac{8670 \text{ eV}}{\text{c}}\\ &=\frac{(8670)(1.60\times10^{-19})}{3\times10^8} \frac{\text{kg m}}{\text{s}}\\ &=4.62\times10^{-24}\ \frac{\text{kg m}}{\text{s}} \end{align} and so we have the wavelength \begin{align} \lambda_n &= \frac{h}{p_n}\\ &=\frac{6.63\times10^{-34}}{4.62\times10^{-24}}\text{ m}\\ &=1.44\times10^{-10}\text{ m}\\ &=.144\text{ nm} \end{align}
For light we have the relationship between energy and frequency of a photon of \begin{align} E = hf \end{align} and a relationship between frequency and wavelength of \begin{align} \lambda f = c \end{align} So if we combine these equations then for a \(1.0\text{MeV}\) photon it follows that \begin{align} \lambda &= \frac{hc}{E}\\ &=\frac{(4.14\times10^{-15})(3\times10^8)}{1.0\times10^6} \text{ m}\\ &=1.24\times10^{-12}\text{ m} \end{align} We now repeat the calculation of the previous problem for a neutron with \(1.0 \text{ MeV}\). First the momentum \begin{align} p_n&=\sqrt{2m_nK_n}\\ &=\sqrt{\frac{2(939 \text{ MeV})(1.0M\text{ eV})}{c^2}}\\ &=\frac{43 \text{ MeV}}{c}\\ &=\frac{(43\times10^6)(1.6\times10^{-19})}{3\times10^8}\frac{\text{kg m}}{\text{s}}\\ &=2.3\times10^{-20}\ \frac{\text{kg m}}{\text{s}} \end{align} so the corresponding wavelength is \begin{align} \lambda_n &= \frac{h}{p_n}\\ &=\frac{6.63\times10^{-34}}{2.3\times10^{-20}}\text{ m}\\ &=2.9\times10^{-14} \text{ m} \end{align} So we conclude that the neutron has a much smaller wavelength.
(Q5B.9 from textbook) A baseball has a mass of \(0.15 \text{ kg}\), and a major-league pitcher can throw a ball with a speed of \(40 \text{ m/s}\) (\(90 \text{ mi/h}\)).
(a) What is the approximate de Broglie wavelength of a beam of baseballs pitched at such a speed?
We first compute the momentum of the baseballs. \begin{align} p &= mv\\ &=(.15)(40)\ \frac{\text{kg m}}{\text{s}}\\ &=6.0\ \frac{\text{kg m}}{\text{s}} \end{align} We then compute the corresponding wavelength \begin{align} \lambda &=\frac{h}{p}\\ &=\frac{6.6\times10^{-34}}{6.0}\text{ m}\\ &=1.1\times10^{-34}\text{ m} \end{align}
(b) Why do we not have to worry much about the wave aspects of a beam of baseballs?
In the first place the baseballs are typically being thrown one at a time, making it less of a continuous wave since the baseball cannot collide with itself. Furthermore, if for instance there was a double slit that a stream of baseballs was being thrown through, our deviation would be approximately \begin{align} \theta = \frac{\lambda}{d} \end{align} where \(d\) is the width of the slits. Given that this slit would be presumablly wide enough for a baseball to fit through, this suggests distances of about \(10^{-34}\) meteres! For reference the width of an atom is about \(10^{-10}\) meters. So this deviation is certianly negligable.
It follows that the momentun of the electron is \begin{align} p&=\frac{m_ec}{137}\\ &=\frac{(9.11\times10^{-31})(3\times10^8)}{137} \frac{\text{kg m}}{\text{s}}\\ &=1.99\times10^{-24}\ \frac{\text{kg m}}{\text{s}} \end{align} Therefore, the corresponding wavelength is \begin{align} \lambda&=\frac{h}{p}\\ &=\frac{6.626\times10^{-34}}{1.99\times10^{-24}}\text{ m}\\ &=3.33\times10^{-10}\text{ m}\\ &=.333\text{ nm} \end{align} with an atomic radius of \(.053\text{ nm}\) we can expect wave behavior to be significant.