Midterm 2 Solutions
- Page ID
- 50497
Problem 1 (2 points each = 20 points total)
MARK the following statements as TRUE or FALSE.
- The Pauli Exclusion Principle states that two or more electrons in a single atom must have the same four quantum numbers.
False
- At absolute zero, all intermolecular and intramolecular motion has ceased.
False
- The square of the wavefunction is measure of the energy of the wavefunction.
False
- The Bohr Principle states that an atom is wavelike only when it has enough mass to exceed the Bohr radius.
False
- Amino acids are never part of the active site of an enzyme?
False
- The number of nodes in a wavefunction is typically correlated to the energy of that wavefunction.
True
- According to Einstein’s Law of photoelectron effect in a primary photoionization process of a metal, each electron is generated by the absorption of a single photon.
True
- Enzymes can make a reaction go faster by reducing DG for the reaction.
False
- Hydrogen-like wavefunctions are derived from solving the Schrodinger equation for a charged nucleus with a single moving electron.
True
- The particle in the box model predicts that the particle has a small probability to be found outside the box as long as the potential walls surrounding the box are of finite height.
True
Problem 2 (10+10=20 points)
2A) What is the energy of a particle with mass of 2.3 x 10-27 kg that is confined in a one-dimensional box of 50 Å length? Assume that the particle is in the third energy level.
\[E_3 = \frac{n^2h^2}{8mL^2} = \frac{9x(6.6x10^{-34}Js)^2}{8x(2.3x10^{-27}kg)(5x10^{-9}m)^2}\]
\[E_3 = 8.5x10^{-24}J\]
2B) What is the wavelength of a photon released during a transition from level three to level two in above system?
\[\Delta E = E_3 - E_2 = \frac{(n_3^2 - n_2^2)h^2}{8mL^2} = \frac{5x(6.6x10^{-34}Js)^2}{8x(2.3x10^{-27}kg)(5x10^{-9}m)^2}\]
\[\Delta E = 4.7x10^{-24}J\]
\[\Delta E = \frac{hc}{\lambda} = \frac{19.8x10^{-26}Jm}{4.7x10^{-24}J} = 4.2 cm\]
Problem 3 (10 + 5 points = 15 total)
3a) The speed of a hydrogen atom is 350 km s-1. If the uncertainty in its momentum is 0.01%, what is the uncertainty in its location? If a hydrogen atom has a diameter of 1.4 Å, how large is this uncertainty compared to its size (i.e., in percent?)
\[(\Delta x)(\Delta p) \geq \frac{\hbar}{2}\]
\[(\Delta x)(350000m/s)(1.673x10^{-27}) \geq 5.3x10^{-35}\]
\[\Delta x \geq 9x10^{-10}m\]
\[\frac{\Delta x}{x} = \frac{9}{1.4} = 6.43 = 643\%\]
Problem 5 (10 points each = 20 points)
B) A reaction that consume CO2 follows Michaelis-menten enzyme kinetics with Km=0.001 M, and the initial concentration of substrate=0.840 M. After 2 seconds, 5% of the substrate has been converted to product[DL1] . How much CO2 will be converted after 10 and 60 seconds?
Answer: 25% after 10 seconds and 100% after 60 seconds
Can either use standard MM equation
\[v_o= k_2[S]_o[E]_o/{[S]_o + K_m}\]
Or recognize that \([S]_o >> Km) so the system is running at V_max (full capacity). Therefor
\[V_max= k_2[E]_o\]
If 5% is consumed in 2 seconds, then 10 second will be 5x5=25%. All of it will be consumed by 60 seconds (by 40 second actually).
3B) When 2 x 10-10 mol is added to a 1-L solution containing its substrate at a concentration one hundred times the Km, it catalyzes the conversion of 75 µmol of substrate into product in three minutes. What is the enzyme’s turnover number (turnover rate)?
\[Kcat = Vmax / [E]_o\]
Because the velocity measured occurs far above Km, it represents Vmax.In three minutes, this amount of enzyme produced 75 µmol of product, equivalent to 25 x 10-6 mol of product per minute.The turnover number is, therefore, (25 x 10-6 mol/min)/(2 x 10-10 mol) = 12.5 x 104 min-1.
Problem 6 (2 + 5 = 8 points total)
Answer the following questions using the diagrams shown for enzymatic catalyzed reactions.
- Which enzyme is providing the least acceleration in relationship to the non-catalyzed reaction? 4
- Which enzyme is providing the greatest acceleration in relationship to the non-catalyzed reaction? 3
- When comparing enzymes 1 and 4, which will be faster? 1
- When comparing enzymes 2 and 3, which will be slower? 2
Problem 7 ( 6 + 3 + 3 = 12 points)
7A) How many electrons in a hydrogen atom can occupy the wavefunctions with the following values of l:
- l =0
1 orbital = 2 electrons
- l =3
7 orbitals = 14 electrons
- l =5
11 orbitals = 22 electrons
7B) What is the probability of finding a particle outside a box (choose either: 0, 1 or in-between)
- with finite potential walls and why?
In between, because the particle can tunneling through walls of finite height.
- with infinite potential walls and why?
Zero since the potential energy is infinite and a particle cannot have infinite energy
7C) What is the probability of finding a particle exactly in the middle of a box (with infinite potential walls) when excited to the n=1 state?
\[\int_{.5L+\epsilon}^{.5L-\epsilon}\frac{2}{L}sin(\frac{\pi x}{L})dx\]
Where after integration we take the limit as \(\epsilon\) goes to zero. For this problem we also accepted as correct:
1. 0 since finding a particle in an exact spot is impossible by the uncertainty principle
2. \[\int_{.5L}^{.5L}\frac{2}{L}sin(\frac{\pi x}{L})dx\]
3. \[\psi ^2 = \frac{2}{L}sin(\frac{\pi}{L}\frac{L}{2}) = \frac{2}{L}\]
Problem 8: 10 points
The diffraction phenomenon can be observed whenever the wavelength is comparable in magnitude to the size of a slit opening. How fast must your instructor, weighing 92 Kg, run to diffract through a door 1 m wide?
\[\lambda = \frac{h}{p} = \frac{h}{mv}\]
\[v = \frac{h}{\lambda m} = \frac{6.6x10^{-34}Js}{92kg} = 7.2x10^{-36}m/s\]