Extra Credit 24

 https://phys.libretexts.org/Bookshel...ves_(Exercise)

Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

#59,  #65, #93

Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

 https://phys.libretexts.org/Bookshel...ves_(Exercise)

#15, #38, #51

Solutions

Question #59

Interstellar space is filled with radiation of wavelength \(970 \mu m\). This radiation is considered to be a remnant of the “big bang.” What is the corresponding blackbody temperature of this radiation?

To solve for the wavelength of maximum energy, we need to use Wien's Displacement Law which relates temperature (\(T\)) to the wavelength (\(\lambda_{max} \)) with a proportionality constant called Wien's displacement constant (\(b = 2.89 \times 10^{-3}m K\)).

\[\lambda_{max} = \dfrac{b}{T}\]

First, rearrange the equation to solve for temperature.

\[T = \dfrac{b}{\lambda_{max}}\]

Now, we can plug in the values for b and \(\lambda\) to evaluate.

\[T = \dfrac{2.89 \times 10^{-3}m K}{970 \mu m} \\ = \dfrac{2.89 \times 10^{-3}m K}{970 \times 10^{-6} m} \\ = 2.97 K \]

The temperature of the radiation is 2.97 K. 

Question #65

Estimate the binding energy of electrons in magnesium, given that the wavelength of 337 nm is the longest wavelength that a photon may have to eject a photoelectron from magnesium photoelectrode.

The binding energy, or work function, (\(\phi\)) is the required energy to free an electron from an atom and it is equal to the threshold frequency (\(\nu_0\)) multiplied by Plank's constant (\(h = 6.626 \times 10^{-34} J \hspace{1mm} s\)). The threshold frequency is the lowest energy light particle needed to satisfy the work function.

\[\phi = \frac{h}{\nu_0} \]

We can change this equation to an expression with the wavelength by using the relationship that frequency is the inverse of the wavelength (\(\lambda\)) times the speed of light (\(c = 3.00 m \hspace{1mm} s^{-1}\)), \(\nu = \frac{c}{\lambda}\).

\[\phi = \dfrac{h  \hspace{1mm}c}{\lambda}\]

Now, we can plug in the values and evaluate.

\[\phi = \dfrac{ (6.626 \times 10^{-34} J \hspace{1mm} s) (3.00 m \hspace{1mm}  s^{-1})}{337 nm} \\ = \dfrac{ (6.626 \times 10^{-34} J \hspace{1mm} s) (3.00 m \hspace{1mm} s^{-1})}{337 \times 10^{-9} m} \\ = 5.90 \times 10^{-19} J\]

The binding energy of electrons in magnesium is about \(5.90 \times 10^{-19} J\). If necessary, we can convert it to eV.

\[ \phi = 5.90 \times 10^{-19} J \times \dfrac{1 eV}{1.60 \times 10^{-19}J} = 3.69 eV\]

Question #93

What is the frequency of the photon absorbed when the hydrogen atom makes the transition from the ground state to the \(n=4\) state?

To solve for the wavelength (\(\lambda\)), we must use the Rydberg Formula

\[\tilde{\nu} = \dfrac{1}{\lambda} = R \left({\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}}\right) \quad n_1 = 1, 2, 3..., n_2 > n_1\]

\tilde{\nu} is the wave number, which is just the inverse of the wavelength. R is the Rydberg constant (\(109,737 cm^{-1}\) or \(2.18 \times 10^{-18} J\)).

We can use the relationship that frequency is the inverse of the wavelength times the speed of light (\(c = 3.00 m \hspace{1mm} s^{-1}\)), \(\nu = \frac{c}{\lambda}\) to rearrange the Rydberg formula to solve for frequency.

\[\dfrac{1}{\lambda} = \dfrac{\nu}{c} = R \left({\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}}\right) \\ \nu =  R \hspace{1mm} c \left({\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}}\right)\]

Now, we can plug in the values and evaluate.

\[ \nu = (109,737 cm^{-1})(3.00 \times 10^8 m \hspace{1mm} s^{-1}) \left({\dfrac{1}{1^2} - \dfrac{1}{4^2}}\right) \\ = (109,737 \times 10^{-2} m^{-1})(3.00 \times 10^8 m \hspace{1mm} s^{-1}) \left({1 - \dfrac{1}{16}}\right) \\ = 3.09 \times 10^{11} s^{-1}\]

The frequency of the photon absorbed when the hydrogen atom makes the transition from the ground state \(n=1\) to the \(n=4\) state is \(3.09 \times 10^{11} s^{-1}\).

Answers

Question #15

Is the photoelectric effect a consequence of the wave character of radiation or is it a consequence of the particle character of radiation? Explain briefly.

The photoelectric effect is a consequence of the wave character of radiation because Young's double split experiment showed that a light source shown through two slits creates a diffraction pattern on the other side, which is characteristic of a wave. If light just had particle characteristics, the light would have shown through the slits directly creating just two lines.

Question #38

Show that Planck’s constant has the dimensions of angular momentum.

Angular momentum (\(L\)) is the product of mass (\(m\)), velocity (\(v\)) and radius (\(r\)).

\[L = m v r\]

If we look at the units for each variable we get the dimensions of the angular momentum as

\[L = kg \times \dfrac{m}{s} \times {m} = kg \hspace{1mm} m^2 \hspace{1mm} s^{-1}\]

Now, lets look at the dimensions of Plank's constant (\(h\)). 

\[h = J \hspace{1mm} s = kg \hspace{1mm} m^2 \hspace{1mm} s^{-2} \times s = kg \hspace{1mm} m^2 \hspace{1mm} s^{-1}\]

Plank's constant has the same dimensions of angular momentum.

Question # 51

Does the Heisenberg uncertainty principle allow a particle to be at rest in a designated region in space?

The Heisenberg uncertainty principle does not allow a particle to be at rest in a designated region in space.