Q#23.6
Energy of DNA Base Pairing, I. (See Exercise 21.23.)
(a) Calculate the electric potential energy of the adenine–thymine bond, using the same combinations of molecules (O-H-N and N-H-N) as in Exercise 21.23.
(b) Compare this energy with the potential energy of the proton–electron pair in the hydrogen atom.
Answer:
The total potential energy is the scalar sum of the individual potential energies of each pair of charges.
For a pair of point charges the electrical potential energy is $U=k\frac{qq'}{r}$
In the O-H-N combination the O$^-$ is 0.170 nm from the H$^+$ and 0.280 nm from the N$^-$. In the N-H-N combination the N$^-$ is 0.190 nm from the H$^+$ and 0.300 nm from the other N$^-$. U is positive for like charges and negative for unlike charges.
(a) O-H-N:
$O^- - H^+$: U = $-(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.170 \times 10^{-9} \ m}=-1.35 \times 10^{-18}$ J
$O^- - N^-$: U = $-(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.280 \times 10^{-9} \ m}=+8.22 \times 10^{-19}$ J
N-H-N:
$N^- - H^+$: U = $-(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.190 \times 10^{-9} \ m}=-1.21 \times 10^{-18}$ J
$N^- - N^-$: U = $-(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.300 \times 10^{-9} \ m}=+7.67 \times 10^{-19}$ J
The total potential energy is
$U_{tot}=-1.35 \times 10^{-18} \ J+8.22 \times 10^{-19} \ J -1.21 \times 10^{-18} \ J+7.67 \times 10^{-19} \ J=-9.71 \times 10^{-19}$ J
(b) In the hydrogen atom the electron is 0.0529 nm from the proton.
U = $-(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.0529 \times 10^{-9} \ m}=-4.35 \times 10^{-18}$ J
The magnitude of the potential energy in the hydrogen atom is about a factor of 4 larger than what it is for the adenine-thymine bond.
Q#23.7
Energy of DNA Base Pairing, II. (See Exercise 21.24.) Calculate the electric potential energy of the guanine– cytosine bond, using the same combinations of molecules (O-H-O, N-H-N, and O-H-N) as in Exercise 21.24.
Answer:
The total potential energy is the scalar sum of the individual potential energies of each pair of charges.
For a pair of point charges the electrical potential energy is $U=k\frac{qq'}{r}$.
In the O-H-O combination the O$^-$ is 0.180 nm from the H$^+$ and 0.290 nm from the other O$^-$. In the N-H-N combination the N$^-$ is 0.190 nm from the H$^+$ and 0.300 nm from the other N$^-$ . In the O-H-N combination the O$^-$ is 0.180 nm from the H$^+$ and 0.290 nm from the N$^-$. U is positive for like charges and negative for unlike charges.
O-H-O: $O^- - H^+$, U = $-1.28 \times 10^{-18}$ J; $O^- - O^-$, U = $+7.93 \times 10^{-19}$ J
N-H-N: $N^- - H^+$, U = $-1.21 \times 10^{-18}$ J; $N^- - N^-$, U = $+7.67 \times 10^{-19}$ J
O-H-N: $O^- - H^+$, U = $-1.28 \times 10^{-18}$ J; $O^- - N^-$, U = $+7.93 \times 10^{-19}$ J
The total potential energy is $-3.77 \times 10^{-18} \ J$ + $2.35 \times 10^{-18} \ J=-1.42 \times 10^{-18}$ J.
For pairs of opposite sign the potential energy is negative and for pairs of the same sign the potential energy is positive. The net electrical potential energy is the algebraic sum of the potential energy of each pair.
Q#23.8
Three equal 1.20 $\mu$C point charges are placed at the corners of an equilateral triangle whose sides are long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
Call the three charges 1, 2 and 3. U = $U_{12} + U_{13} + U_{23}$.
$U_{12} = U_{13} = U_{23}$ because the charges are equal and each pair of charges has the same separation, 0.500 m.
U = $\frac{3kq^2}{r}=\frac{3k(1.2 \times 10^{-6} \ C)}{0.500 \ m}=0.078 \ J$
When the three charges are brought in from infinity to the corners of the triangle, the repulsive electrical forces between each pair of charges do negative work and electrical potential energy is stored.
Q#23.9
Two protons are released from rest when they are 0.750 nm apart.
(a) What is the maximum speed they will reach? When does this speed occur?
(b) What is the maximum acceleration they will achieve? When does this acceleration occur?
Answer:
The protons repel each other and therefore accelerate away from one another. As they get farther and farther away, their kinetic energy gets greater and greater but their acceleration keeps decreasing. Conservation of energy and Newton’s laws apply to these protons.
Let a be the point when they are 0.750 nm apart and b be the point when they are very far apart. A proton has charge +e and mass 1.67 $\times 10^{-27}$ kg. As they move apart the protons have equal kinetic energies and speeds. Their potential energy is
U = $ke^2/r^2$ and K = $\frac{1}{2}mv^2$. $K_a+U_a=K_b+U_b$
(a) They have maximum speed when they are far apart and all their initial electrical potential energy has been converted to kinetic energy. $K_a+U_a=K_b+U_b$
$K_a=0$, $U_b=0$, so
$K_b=U_a=ke^2/r_a=(8.988 \times 10^9 \ N.m^2/C^2)\frac{(1.60 \times 10^{-19} \ C)^2}{0.750 \times 10^{-9} \ m}=3.07 \times 10^{-19} \ J$
$K_b=\frac{1}{2}mv_b^2+\frac{1}{2}mv_b^2=mv_b^2$
$v_b=\sqrt{\frac{K_b}{m}}=\sqrt{\frac{3.07 \times 10^{-19} \ J}{1.67 \times 10^{-19} \ kg}}=1.36 \times 10^4$ m/s
(b) Their acceleration is largest when the force between them is largest and this occurs at r = . 0 750 nm, when they are closest.
$F=\frac{ke^2}{r^2}=(8.988 \times 10^9 \ N.m^2/C^2)\left(\frac{1.60 \times 10^{-19} \ C}{0.750 \times 10^{-9} \ m}\right)^2=4.09 \times 10^{-10}$ N
$a = \frac{F}{m}=\frac{4.09 \times 10^{-10} \ N}{1.67 \times 10^{-27} \ kg}=2.45 \times 10^{17} \ m/s^2$
The acceleration of the protons decreases as they move farther apart, but the force between them is repulsive so they continue to increase their speeds and hence their kinetic energies.
Q#23.10
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?
Answer:
The work done on the alpha particle is equal to the difference in its potential energy when it is moved from the midpoint of the square to the midpoint of one of the sides.
We apply the formula $W_{a→b}=U_a-U_b$. In this case, a is the center of the square and b is the midpoint of one of the sides. Therefore $W_{center→site}=U_{center}-U_{site}$ is the work done by the Coulomb force. There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single alpha-electron pair.
At the center of the square, the alpha particle is a distance $r_1=\sqrt{50}$ nm from each electron. At the midpoint of the side, the alpha is a distance $r_2$ = 5.00 nm from the two nearest electrons and a distance $r_3=\sqrt{124}$ nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of two point charges,
U = $\frac{kqq_0}{r^2}$
the total work done by the Coulomb force is
$W_{center→site}=U_{center}-U_{site}$
$W_{center→site}=4k\frac{q_{\alpha}q_e}{r_1}-\left(2k\frac{q_{\alpha}q_e}{r_2}+2k\frac{q_{\alpha}q_e}{r_3}\right)$
Substituting $q_e=-e$ and $q_{\alpha}=2e$ and simplifying gives
$W_{center→site}=-4ke^2\left[\frac{2}{r_1}-\left(\frac{1}{r_2}+\frac{1}{r_3}\right)\right]$
Substituting the numerical values into the equation for the work gives
$W_{center→site}=-4(8.988 \times 10^{9} \ N.m^2/C^2)(1.60 \times 10^{-19} \ C)\times$
$\left[\frac{2}{\sqrt{50} \ nm}-\left(\frac{1}{5.00 \ nm}+\frac{1}{\sqrt{125} \ nm}\right)\right]=6.08 \times 10^{-21}$ J
Since the work done by the Coulomb force is positive, the system has more potential energy with the alpha particle at the center of the square than it does with it at the midpoint of a side. To move the alpha particle to the midpoint of a side and leave it there at rest an external force must do $-6.08 \times 10^{-21}$ J of work.
Q#23.11
Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d. Two of the point charges are identical and have charge If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?
Answer:
Apply Eq.
$W_{a→b}=U_a-U_b=-\Delta U$.
The net work to bring the charges in from infinity is equal to the change in potential energy. The total potential energy is the sum of the potential energies of each pair of charges, calculated from Eq.
U = $\frac{kqq_0}{r^2}$.
Let 1 be where all the charges are infinitely far apart. Let 2 be where the charges are at the corners of the triangle, as shown in Figure 23.11.
Let $q_c$ be the third, unknown charge.
W = $-\Delta U=-(U_2-U_1)$ where W is the work done by the Coulomb force.$U_1=0$
$U_2=U_{ab}+U_{bc}+U_{ac}$
$U_2=\frac{1}{4\pi \epsilon_0}d(q^2+2qq_c)$
Want W = 0, so W = $-\Delta U=-(U_2-U_1)$ gives 0 = $-U_2$
$0=\frac{1}{4\pi \epsilon_0}d(q^2+2qq_c)$
$0=q^2+2qq_c$ and $q_c=-q/2$
The potential energy for the two charges q is positive and for each q with $q_c$ it is negative. There are two of the q, $q_c$ terms so must have $q_c<q$.
Q#23.12
Starting from a separation of several meters, two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1000 m/s measured relative to the earth. Find the maximum electrical force that these protons will exert on each other.
Answer:
Use conservation of energy $K_a+U_a=K_b+U_b$ to find the distance of closest approach $r_b$.
The maximum force is at the distance of closest approach, $F=k\frac{|q_1||q_2|}{r_b^2}$
$K_b$ = Initially the two protons are far apart, so $U_a$ = 0. A proton has mass 1.67 $\times 10^{-27}$ kg and charge q = +e = 1.60 $\times 10^{-19}$ C.
$K_a=U_b$
$2.\frac{1}{2}mv_a^2=k\frac{q_1q_2}{r_b}$
$mv_a^2=k\frac{e^2}{r_b^2}$ and
$r_b=\frac{ke^2}{mv_a^2}$
$r_b=\frac{(8.99 \times 10^9 \ N.m^2/C^2)(1.60 \times 10^{-19} \ C)^2}{(1.67 \times 10^{-27} \ kg)(1.00 \times 10^6 \ m/s)^2}=1.38 \times 10^{-13} \ m$
$F=k\frac{e^2}{r^2}=\frac{(8.99 \times 10^9 \ N.m^2/ C^2)(1.60 \times 10^{-19} \ C)^2}{(1.38 \times 10^{-13} \ m)^2}=0.012 \ N$
The acceleration a = F/m = of each proton produced by this force is extremely large.
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