Extra Credit 32

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

https://phys.libretexts.org/Bookshel...cs_(Exercises)

#34, #86, #89

The Heisenberg uncertainty principle is defined as: 

\begin{equation}
\Delta y*\Delta p \geq \frac{\hbar}{2}
\end{equation}

(a) where

\begin{equation}
p=m*\Delta v
\end{equation}

so

\begin{equation}
\Delta y*m*\Delta v \geq \frac{\hbar}{2}
\end{equation}

Solving for the y component of velocity

\begin{equation}
\Delta v = \frac{\hbar}{2*\Delta y*m}
\end{equation}

plug in known values:
\hbar = 1.055E-34 J*s
m = 1.673E-27 kg
\Delta y = 2.0E-12 m

\begin{equation}
\Delta v= \frac{1.055E-34 J*s}{2* 1.673E-27 kg * 2.0E-12 m}
\end{equation}

\Delta v = 1.577E4 m/s

(b) If the uncertainty in x position) 
If the uncertainty of the position is zero then the uncertainty in the velocity is at minimum 'c' or the speed of light.

 

 

(a) The Energy of the Third Harmonic Oscillator is given by

\begin{equation}
E_3= \frac{7}{2}*\hbar\sqrt{\frac{k}{m}}
\end{equation}

The Energy of the Second Harmonic Oscillator is given by
 

\begin{equation}
E_2= \frac{5}{2}*\hbar\sqrt{\frac{k}{m}}
\end{equation}

Subtracting the two:

\begin{equation}
E_3 - E_2 = \frac{h}{2\pi}*\sqrt{\frac{k}{m}} = \frac{h*c}{\lambda}
\end{equation}

To find the wavelength, separate \lambda to one side:
\begin{equation}
\lambda = 2c\pi*\sqrt{\frac{k}{m}}
\end{equation}

where h is the Planck constant
m is the mass given 5.6E-26 kg
k is the force constant 12 N/m
c is the speed of light

Plugging each constant in

\begin{equation}
\lambda = 2*2.998E8*\pi*\sqrt{\frac{12 N/m}{5.6E-26 kg}}
\end{equation}

the wavelength is = 1.29E-4 m

 

(b) Find the ground state

Ground state equation is given by
\begin{equation}
\frac{h}{4\pi}*\sqrt{\frac{k}{m}} + E_0
\end{equation}

Plug in all of the values as the above to solve for E_0

E_0 = 7.72E -22 Joules

 

The equation for the transional energy of a particle in a box is 

\begin{equation}
E= n^2*\frac{\pi^2\hbar^2}{2mL^2}
\end{equation}

(a) Energy for the particle of the box from first state to ground state

Substitute 2 and 1 into the values for n in the equation above

\begin{equation}
E_2= 2^2*\frac{\pi^2\hbar^2}{2mL^2} = E
\end{equation}

\begin{equation}
E_1= 1^2*\frac{\pi^2\hbar^2}{2mL^2} = E
\end{equation}

\begin{equation}
E_2-E_1= 3*\frac{\pi^2\hbar^2}{2mL^2} = E
\end{equation}

To find the wavelength of a proton
\begin{equation}
E= h*\frac{c}{\lambda}
\end{equation}

Substitute in

\begin{equation}
h*\frac{c}{\lambda} = 3*\frac{\pi^2\hbar^2}{2mL^2}
\end{equation}

solving to find the wavelength

\begin{equation}
\lambda = \frac{hc2mL^2}{3\hbar^2\pi^2}
\end{equation}

Plug in the constants where:
h is the Planck constant 6.626E-34
hbar is equal to 1.055E-34 J*2
m is the mass given 9.1E-31 kg
L is the length of the box 4.18nm 
c is the speed of light

the wavelength is = 19.2E-6 meters

(b) from the second excited state to the first

\begin{equation}
E_2= 2^2*\frac{\pi^2\hbar^2}{2mL^2} = E
\end{equation}

\begin{equation}
E_3= 3^2*\frac{\pi^2\hbar^2}{2mL^2} = E
\end{equation}

Rearraging for \lambda 

\begin{equation}
\lambda = \frac{hc2mL^2}{5\hbar^2\pi^2}
\end{equation}

Plugging in all of the same constants as above, the wavelength is 

the wavelength is = 7.67E-6 meters

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...cs_(Exercises)

#3, #8, #13

 

#3) 

Question: What kind of physical quantity does a wave function of an electron represent?

Answer: the wave function of an electron has no physical significance. However, if you square the wave function you can get a probability density that can be used to determine where an electron can be located within an atom. 

 

#8)

Question:  Can we measure the energy of a free localized particle with complete precision?

Answer: You cannot measure the energy of a free localized particle because the Heisenberg Uncertainty Principle states that when we have a 100% certainty in one aspect, we would have infinitely high uncertainty in the value for t at the point when we measure the energy. The equation

\begin{equation}
\Delta*E*\Delta*t \geq \frac{h}{2\pi}
\end{equation}

 

#13) 

Question: Suppose a wave function is discontinuous at some point. Can this function represent the quantum state of some physical particle? Why? Why not?

Answer: A discontinuous wave function cannot represent a quantum state because, by definition, a wave function must be a continuous function. There will never be a point in space in which there is no matter as defined by the wave function. Also, at this point, there will be an undefined probability of finding an electron which is not possible. There can be 0 probability, or a range of probabilities but not an undefined probability. Therefore, it is not possible to represent the quantum state of this particle at a discontinuous point.