Bohr radius =
Energy of a matter wave (electron) can only exist in discrete states. The energy associated with the quantum states of the matter wave are given by:
\[ E_{n} = ( \frac{h^{2}}{8 m L^{2}} ) n^{2} \]
Where:
An electron can jump from a quantum state to another only if its energy change is
\[ \Delta E = E_{high} - E_{low} \]
Where:
If the specified energy change is done by an absorbed/emitted photon, then we know this energy must be equal to \(\Delta E\):
\[ hf = \frac{hc}{\lambda} = \Delta E = E_{high} - E_{low} \]
Where:
Wave function for an electron in an infinite, one-dimensional particle well with length \(L\) along an \(x\) axis are given by:
\[ \psi_{n} (x) = \sqrt{\frac{2}{L}} sin ( \frac{n \pi}{L} x ) \]
Where:
The product \(\psi_{n}^{2} (x) dx\) is the probability that the electron will be detected in the interval between coordinates \(x\) and \(x + dx\).
If the probability density of an electron is integrated over the entire \(x\) axis, the total probability must be 1:
\[ \int_{-\infty}^{\infty} \psi_{n}^{2} (x) dx = 1 \]
Quantized energy of a two-dimensional infinite potential well is given by the two independent quantizations of the axes:
\[ E_{nx,ny} = \frac{h^{2}}{8m} ( \frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}} ) \]
The wave function for an electron in a two-dimensional well are given by:
\[ \psi_{nx,ny} = \sqrt{\frac{2}{L_{x}}} sin ( \frac{n_{x} \pi}{L_{x}} x ) \sqrt{\frac{2}{L_{y}}} sin ( \frac{n_{y} \pi}{L_{y}} y ) \]
Bohr model assumption regarding angular momentum \(L\) is quantized, and given by:
\[ L = n \hbar \]
Where:
Applying Schrodinger’s equation gives us the correct values of \(L\) and the quantized energies:
\[ E_{n} = - \frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \frac{1}{n^{2}} = - \frac{13.61\text{ eV}}{n^{2}} \]
Where:
If an electron changes energy levels by absorbing/emitting a photon, then the wavelength of light can be represented as:
\[ \frac{1}{\lambda} = R ( \frac{1}{n^{2}_{low}} - \frac{1}{n^{2}_{high}} ) \]
Where:
\(R\) is the Rydberg constant:
\[ R = \frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c} = 1.097373 \times 10^{7} \text{m}^{-1} \]
The radial probability density \(P(r)\) for a state of the hydrogen atom is defined so that \(P(r) dr\) is the probability that the electron will be detected in the space between two spherical shells of radii \(r\) and \(r + dr\) that are centered on the nucleus.
Normalization requires that:
\[ \int_{0}^{\infty} P(r) dr = 1 \]
The probability that the electron will be detected between any two given radii \(r_{1}\) and \(r_{2}\) is:
\[ \int_{r_{1}}^{r_{2}} P(r) dr \]
This question is regarding energies of a trapped electron.
What is the ground-state energy of
- an electron
- a proton if each is trapped in a one-dimensional infinite potential well that is 200 pm wide?
For this question, we use this:
\[ E_{n} = ( \frac{h^{2}}{8 m L^{2}} ) n^{2} \]
The ground state corresponds to the quantum number 1.
Using the mass of an electron…
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (200 \times 10^{-12})^{2}} ) 1^{2} \]
Thus, we get \(E_{n}\) to be \(1.5 \times 10^{-18}\) J as our answer for the ground-state energy of an electron.
Now, we use the mass of a proton:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (1.67 \times 10^{-27}) * (200 \times 10^{-12})^{2}} ) 1^{2} \]
Thus, we get \(E_{n}\) to be \(8.2 \times 10^{-22}\) J as our answer for the ground-state energy of an proton.
This question is regarding energies of a trapped electron.
An electron, trapped in a one-dimensional infinite potential well 250 pm wide, is in its ground state. How much energy must it absorb if it is to jump up to the state with \(n = 4\)?
First, let us calculate the ground-state energy \(n = 1\) of an electron for a one-dimensional infinite potential well of 250 pm…:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 1^{2} \]
We find \(E_{n}\) at the ground state to be \(9.6 \times 10^{-19}\) J.
Then, we calculate the energy for the \(n = 4\) energy level:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 4^{2} \]
We find \(E_{n}\) at the third excitation level (\(n = 4\)) to be \(1.5 \times 10^{-17}\) J.
Then, we take the difference to find the amount of energy needed to be absorbed for the quantum jump:
\[ \Delta E = E_{high} - E_{low} = 1.5 \times 10^{-17} - 9.6 \times 10^{-19} = 1.4 \times 10^{-17} \]
Thus, our answer is \(1.4 \times 10^{-17}\) J.
This question is regarding energies of a trapped electron.
Consider an atomic nucleus to be equivalent to a one-dimensional infinite potential well with \(L = 1.4 \times 10^{-14}\) m, a typical nuclear diameter. What would be the ground-state energy of an electron if it were trapped in such a potential well? (Note: Nuclei do not contain electrons.)
Find energy of electron trapped in one-dimensional infinite potential well of \(1.4 \times 10^{-14}\) m…:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (1.4 \times 10^{-14})^{2}} ) 1^{2} \]
Thus, our answer is \(3.1 \times 10^{-10}\) J, or \(1.9 \times 10^{9}\) eV.
This question is regarding energies of a trapped electron.
An electron is trapped in a one-dimensional infinite well of width 250 pm and is in its ground state. What are the
- longest
- second longest
- third longest
wavelengths of light that can excite the electron from the ground state via a single photon absorption?
We know that the energy of a photon is given by:
\[ E = hf = \frac{hc}{\lambda} \]
Thus, the wavelength that we can use to excite the electron to the first excitation level is going to be the longest possible, as the energy is quantized.
So, let us first find the energy needed to excite the electron from the ground state to the first excitation level.
The ground state energy is given by:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 1^{2} \]
This comes out to be \(9.6 \times 10^{-19}\) J.
Now, let us find the energy for the first excitation level (\(n = 2\)):
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 2^{2} \]
This comes out to be \(3.9 \times 10^{-18}\) J.
The energy needed to excite the electron from the ground state to the first excitation level is then \(3.9 \times 10^{-18} - 9.6 \times 10^{-19} = 2.9 \times 10^{-18}\) J. Now, if we were to assume this energy is from an absorbed photon, the energy can be represented as:
\[ 2.9 \times 10^{-18} = \frac{hc}{\lambda} \]
\[ 2.9 \times 10^{-18} = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^{8})}{\lambda} \]
Solve for \(\lambda\):
\[ \lambda = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^{8})}{2.9 \times 10^{-18}} \]
Thus, our answer for the first part is \(\lambda = 6.9 \times 10^{-8}\) m.
Now, we repeat this for the second and third excitation levels respectively to the next two parts:
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 3^{2} \]
We find \(E_{n}\) to be \(8.7 \times 10^{-18}\) J.
Then we find the difference between the second excitation level and the ground state to be \(7.7 \times 10^{-18}\) J (this is the energy needed to excite the electron to energy level of quantum number \(n = 3\)).
Finally, we can solve for \(\lambda\):
\[ \lambda = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^{8})}{7.7 \times 10^{-18}} \]
Thus, our answer for the second part is \(\lambda = 2.6 \times 10^{-8}\) m.
\[ E_{n} = ( \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31}) * (250 \times 10^{-12})^{2}} ) 4^{2} \]
We find \(E_{n}\) to be \(1.5 \times 10^{-17}\) J.
Then we find the difference between the third excitation level and the ground state to be \(1.4 \times 10^{-18}\) J (this is the energy needed to excite the electron to energy level of quantum number \(n = 4\)).
Finally, we can solve for \(\lambda\):
\[ \lambda = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^{8})}{1.4 \times 10^{-17}} \]
Thus, our answer for the second part is \(\lambda = 1.4 \times 10^{-8}\) m.
This question is regarding wave functions of a trapped electron.
A one-dimensional infinite well of length 200 pm contains an electron in its third excited state. We position an electron-detector probe of width 2.00 pm so that it is centered on a point of maximum probability density.
- What is the probability of detection by the probe?
- If we insert the probe as described 1000 times, how many times should we expect the electron to materialize on the end of the probe (and thus be detected)?
The probability the electron will be detected is given by:
\[ \psi^{2}_{n} (x) dx = ( \sqrt{\frac{2}{L}} sin ( \frac{n \pi}{L} x ) )^{2} dx \]
We know \(dx = 2 \times 10^{-12}\) m, and that the \(sin(\frac{n \pi}{L} x)\) expression must evaluate to 1 since the probe is centered on a point of maximum probability density:
\[ \psi^{2}_{n} (x) dx = ( \sqrt{\frac{2}{L}} * 1 ) * 2 \times 10^{-12} \]
\[ \psi^{2}_{n} (x) dx = \frac{2}{L} * 2 \times 10^{-12} \]
\(L\) is \(200 \times 10^{-12}\) m:
\[ \psi^{2}_{n} (x) dx = \frac{2}{200 \times 10^{-12}} * 2 \times 10^{-12} \]
This comes out to be a probability of 0.020, our answer to the first part.
If the probe is inserted 1000 times, then we should expect the electron to be detected 20 of the times (\(1000 * 0.020 = 20\)), which is our answer to the second part.
This question is regarding wave functions of a trapped electron.
A particle is confined to the one-dimensional infinite potential well of Fig. 39-2. If the particle is in its ground state, what is its probability of detection between
- x = 0 and x = 0.25L
- x = 0.75L and x = L, and
- x = 0.25L and x = 0.75L?
We need to integrate over the probability density function to approach this question.
\[ \int_{0}^{0.25L} \psi_{n}^{2} (x) dx = \int_{0}^{0.25L} \frac{2}{L} sin^{2} ( \frac{n \pi}{L} x ) dx \]
Let us do the indefinite integral first. Take out the constant factors. We also know that the particle is in the ground state, so \(n = 1\).
\[ \frac{2}{L} \int sin^{2} ( \frac{\pi}{L} x ) dx \]
Now, integrate.
Use double angle sine squared identity (\(sin^{2} x = \frac{1}{2} (1 - cos 2x)\))
\[ \frac{2}{L} \int \frac{1}{2} (1 - cos(2 \frac{\pi}{L} x)) dx \]
\[ \frac{2}{L} \frac{1}{2} ( \int 1 dx - \int cos(2 \frac{\pi}{L} x) dx ) \]
\[ \frac{1}{L} ( x - \int cos(2 \frac{\pi}{L} x) dx) \]
Say \(u = 2 \frac{\pi}{L} x\), then:
\[ \frac{du}{dx} = 2 \frac{\pi}{L} \]
\[ dx = \frac{du}{2 \frac{\pi}{L}} = \frac{L}{2 \pi} du \]
We can now solve
\[ \int cos(2 \frac{\pi}{L} x) dx \]
as:
\[ \frac{L}{2 \pi} \int cos(u) du \]
Which is:
\[ \frac{L}{2 \pi} sin u = \frac{L}{2 \pi} sin ( 2 \frac{\pi}{L} x ) \]
Now, we can substitute that into the main expression:
\[ F(x) = \frac{1}{L} ( x - \frac{L}{2 \pi} sin ( 2 \frac{\pi}{L} x )) \]
Considering the problem gives us inputs in terms of \(L\), let us solve this generically. To be specific, assume the lower and upper bounds can be represented respectively as \(aL\) and \(bL\), where \(a\) and \(b\) are constants that are between 0 and 1 inclusive. The fundamental theorem of calculus states our answer to the definite integral:
\[ \int_{aL}^{bL} \psi_{n}^{2} (x) dx = F(bL) - F(aL) \]
Let us first evaluate \(F(bL)\):
\[ F(bL) = \frac{1}{L} ( bL - \frac{L}{2 \pi} sin ( 2 \frac{\pi}{L} bL )) \]
\[ F(bL) = (b - \frac{1}{2 \pi} sin ( 2 b \pi)) \]
Obviously, \(F(aL)\) will then be:
\[ F(aL) = (a - \frac{1}{2 \pi} sin ( 2 a \pi)) \]
So, \(F(bL) - F(aL)\) is:
\[ F(bL) - F(aL) = (b - \frac{1}{2 \pi} sin ( 2 b \pi)) - (a - \frac{1}{2 \pi} sin ( 2 a \pi)) = b - a - \frac{1}{2 \pi} sin ( 2 b \pi) + \frac{1}{2 \pi} sin ( 2 a \pi) \]
For the first part we can solve using \(a = 0\) and \(b = 0.25\):
\[ 0.25 - \frac{1}{2 \pi} sin (\frac{\pi}{2}) + \frac{1}{2 \pi} sin (0) \]
\[ 0.25 - \frac{1}{2 \pi} + 0 \]
Thus, 0.09 is our answer for the first part.
For the second part we can solve using \(a = 0.75\) and \(b = 1\):
\[ 0.25 - \frac{1}{2 \pi} sin ( 2 \pi) + \frac{1}{2 \pi} sin ( \frac{3}{2} \pi) \]
Note, \(sin(\frac{3}{2} \pi) = sin(\frac{1}{2} \pi) = 1\)
\[ 0.25 + \frac{1}{2 \pi} \]
Thus, 0.09 is our answer for the second part as well.
For the third part, notice that the range given is complementary to the sum of the two ranges from the previous two parts. Since \(\psi^{2}\) is the probability density function, we know:
\[ \int_{-\infty}^{\infty} \psi_{n}^{2} (x) dx = 1 \]
So, \(1 - 0.09 - 0.09 = 0.82\). \(0.82\) is our answer for the third part.
This question is regarding electrons in finite well traps.
An electron in the n = 2 state in the finite potential well of Fig. 39-7 absorbs 400 eV of energy from an external source. Using the energy-level diagram of Fig. 39-9, determine the electron’s kinetic energy after this absorption, assuming that the electron moves to a position for which x > L.
The energy at \(n = 2\) is 106 eV. So if the electron absorbs 400 eV of energy, it goes into the nonquantized region of the figure, since the well depth is \(450\) eV. Since the well depth tells us the minimum energy needed for the confined electron to become free, the kinetic energy is thus \((106 + 400) - 450 = 56\), so 56 eV is our answer.
This question is regarding electrons in finite well traps.
Figure 39-28a shows the energy-level diagram for a finite, one-dimensional energy well that contains an electron. The non-quantized region begins at \(E_{4}\) = 450.0 eV. Figure 39-28b gives the absorption spectrum of the electron when it is in the ground state it can absorb at the indicated wavelengths: \(\lambda_{a}\) = 14.588 nm and \(\lambda_{b}\) = 4.8437 nm and for any wavelength less than \(\lambda_{c}\) = 2.9108 nm. What is the energy of the first excited state?
First, we can determine that if a photon of wavelength \(\lambda_{c}\) is absorbed, this will excite the electron from the ground state to \(E_{4}\). This is because it is the shortest wavelength, so it has the highest energy value (question also states that the wavelength can be less than \(\lambda_{c}\) because of the nonquantized region). Thus, we can find the energy of a photon that has wavelength \(\lambda_{c}\):
\[ E = \frac{hc}{\lambda} = \frac{1240}{2.9108} = 426\text{ eV} \]
This energy represents the energy needed to excite the electron from \(E_{1}\) to \(E_{4}\) (\(E_{4} - E_{1}\)). Since we want the energy of the first excited state this corresponds to \(E_{2}\). We can represent \(E_{2}\) as:
\[ (E_{4} - (E_{4} - E_{1})) + (E_{2} - E_{1}) \]
Notice that \((E_{4} - (E_{4} - E_{1}))\) is just a convenient way of writing \(E_{1}\) in terms of \(E_{4}\) and \(E_{4} - E_{1}\), since this is the information we have.
We already have \((E_{4} - E_{1})\) and \(E_{4}\), so let us find \((E_{2} - E_{1})\). The relevant wavelength here is \(\lambda_{a} = 14.588\text{ nm}\) because we choose the longest wavelength since we are doing a singular quantum jump:
\[ E = \frac{hc}{\lambda} = \frac{1240}{14.588} = 85\text{ eV} \]
So, this represents \(E_{2} - E_{1}\), now we can just evaluate the previous expression:
\[ (E_{4} - (E_{4} - E_{1})) + (E_{2} - E_{1}) \]
\[ (450 - 426) + (85) \]
This comes out to be \(109.0\) eV, our answer.
This question is regarding electrons in finite well traps.
- Show that for the region x > L in the finite potential well of Fig. 39-7, \(\psi (x) = D e^{2kx}\) is a solution of Schrodinger’s equation in its one-dimensional form, where D is a constant and k is positive.
- On what basis do we find this mathematically acceptable solution to be physically unacceptable?
First, let’s take a look at Schrodinger’s equation:
\[ \frac{d^{2} \psi}{dx^{2}} + \frac{8 \pi^{2} m}{h^{2}} ( E - U_{0} ) \psi = 0 \]
Let us find the second derivative of the wave function, \(\frac{d^{2} \psi}{dx^{2}}\). This is going to be \(2k * 2k * D e^{2kx} = 4k^{2} D e^{2kx}\). Then, we can substitute:
\[ 4k^{2} \psi + \frac{8 \pi^{2} m}{h^{2}} ( E - U_{0} ) \psi = 0 \]
\[ 4k^{2} + \frac{8 \pi^{2} m}{h^{2}} ( E - U_{0} ) = 0 \]
Solve for \(k\):
\[ k^{2} = \frac{-8 \pi^{2} m}{-4 h^{2}} ( E - U_{0} ) \]
\[ k = \sqrt{\frac{2 \pi^{2} m}{h^{2}} ( E - U_{0} )} \]
So \(\psi (x) = D e^{2kx}\) is a solution to Schrodinger’s one-dimensional equation.
We know that this is mathematically acceptable because \(\frac{2 \pi^{2} m}{h^{2}}\) is a positive value, and \(U_{0} - E\) is positive since the well height (\(U_{0}\)) is greater than the energy of the particle \(E\) when \(x > L\). This means that the term inside of the square root is positive, so we do not have any imaginary solutions.
This is not physically acceptable though because \(\int_{-\infty}^{\infty} \psi_{n}^{2} (x) dx \neq 1\). Given that \(k\) is positive, the area under the curve is going to keep growing when approaching to infinity, so this integral will not evaluate to 1.
This question is regarding two and three dimensional electron traps.
An electron is contained in the rectangular box of Fig. 39-14, with widths \(L_{x}\) = 800 pm, \(L_{y}\) = 1600 pm, and \(L_{z}\) = 390 pm. What is the electron’s ground-state energy?
We can find the ground-state energy using:
\[ E_{nx,ny,nz} = \frac{h^{2}}{8m} ( \frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}} + \frac{n_{z}^{2}}{L_{z}^{2}}) \]
\[ E_{nx,ny,nz} = \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31})} ( \frac{1}{(800 \times 10^{-12})^{2}} + \frac{1}{(1600 \times 10^{-12})^{2}} + \frac{1}{(390 \times 10^{-12})^{2}}) \]
This comes out to be \(5.14 \times 10^{-19}\) J, or \(3.21\) eV.
This question is regarding two and three dimensional electron traps.
Figure 39-30 shows a two-dimensional, infinite-potential well lying in an \(xy\) plane that contains an electron. We probe for the electron along a line that bisects \(L_{x}\) and find three points at which the detection probability is maximum. Those points are separated by 2.00 nm. Then we probe along a line that bisects \(L_{y}\) and find five points at which the detection probability is maximum. Those points are separated by 3.00 nm. What is the energy of the electron?
Since we observe 3 probability maximums for the x axis and 5 probability maximums for the y axis, the quantum numbers \(n_{x}\) and \(n_{y}\) are \(3\) and \(5\) respectively. We can also find \(L_{x}\) and \(L_{y}\), which are \(3 * (2 \times 10^{-9})\) m and \(5 * (3 \times 10^{-9})\)
We can then find the energy using:
\[ E_{nx,ny} = \frac{h^{2}}{8m} ( \frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}} ) \]
\[ E_{nx,ny} = \frac{(6.626 \times 10^{-34})^{2}}{8 * (9.11 \times 10^{-31})} ( \frac{(3)^{2}}{(3 * (2 \times 10^{-9}))^{2}} + \frac{(5)^{2}}{(5 * (3 \times 10^{-9}))^{2}} ) \]
Thus, our answer is \(E_{nx,ny} = 2.18 \times 10^{-20}\) J, or \(0.136\) eV.
This question is regarding the hydrogen atom.
What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?
First, some definitions:
The quantized energy of the hydrogen atom is inversely proportional to \(n^{2}\), so for \(n = 2\) the energy is \(\frac{1}{4}\) of the energy for \(n = 1\). Energy can be expressed as:
\[ E = \frac{hc}{\lambda} \]
We can see that wavelength for first excited state \(n = 2\) is going to be four times greater than the wavelength for the ground state \(n = 1\), since \(\lambda\) is inversely proportional to the energy.
Therefore, the ratio is going to be 4, our answer.
This question is regarding the hydrogen atom.
What are the
- energy
- magnitude of the momentum
- wavelength
of the photon emitted when a hydrogen atom undergoes a transition from a state with \(n = 3\) to a state with \(n = 1\)?
Energy is given by:
\[ \frac{-13.61}{3^{2}} - \frac{-13.61}{1^{2}} \]
This comes out to be 12.1 eV, our answer.
Momentum is given by:
\[ p = \frac{E}{c} = \frac{12.1\text{ eV} * 1.602 \times 10^{-19}}{3 \times 10^{8}} \]
This comes out to be \(6.45 \times 10^{-27}\) kg m/s, our answer.
Wavelength of the emitted photon is given by:
\[ \lambda = \frac{hc}{E} = \frac{1240}{12.1} \]
This comes out to be \(102\) nm, our answer.
This question is regarding the hydrogen atom.
In the ground state of the hydrogen atom, the electron has a total energy of -13.6 eV. What are
- its kinetic energy
- its potential energy
if the electron is one Bohr radius from the central nucleus?
First, we know the potential energy is given by:
\[ U(r) = \frac{-e^{2}}{4 \pi \varepsilon_{0} r} \]
In this case, \(r = a\), so:
\[ U(r) = \frac{-(1.602 \times 10^{-19})^{2}}{4 \pi \varepsilon_{0} * (5.292 \times 10^{-11})} \]
This comes out to be \(-4.36 \times 10^{-18}\) J, or \(-27.2\) eV, our answer for the second part.
Now we know the total energy is composed of the kinetic and potential energy, so we can find the kinetic energy:
\[ E = U + K \]
\[ K = E - U = -13.6 - (-27.2) = 13.6 \]
Thus, \(13.6\) eV is our answer for the first part.