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(a)
To Calculate: The net force between the thymine and adenine.
(a)
Answer to Problem 51P
The net force is 4.6×10-10N .
Explanation of Solution
Given:
The given helix-shaped DNA molecule
The net average charge on hydrogen and nitrogen atom is 0.2e , on carbon and oxygen atom is 0.40e and the separation distance between each molecule is 1.0×10-10m .
Formula used:
The net force formula is given as,
F= kq1q2r2
Where, k is the proportionally constant, q1and q2 are testing charges and r is the separation distance between the testing charges.
The net force is calculated as,
FC.G= 2FO−H+2FO−N+FH−N+FN−N
Calculation:
Draw the figure for the thymine and adenine.
Here, in this bonds which are responsible are NHO and NHN.
For NHO bonds:
For the O-H attraction, FO−H= k(0.4e)×(0.2e)(1.80×10−10m)2=0.08ke2(1.90×10−10m)2
For the O-N repulsion, FO−N= k(0.4e)×(0.2e)(2.80×10-10m)2=0.08ke2(2.80×10-10m)2
For NHN bond:
For the H-N attraction FH−N= k(0.2e)×(0.2e)(2.00×10−10m)2=0.04ke2(2.00×10−10m)2
For the N-N repulsion, FN−N= k(0.2e)×(0.2e)(3.00×10−10m)2=0.04ke2(3.00×10−10m)2
The net force is given as,
FC.G= FO−H−FO−N−FN−N+FH−N = (0.08ke2(1.80×10−10m)2−0.08ke2(2.80×10−10m)2−0.08ke2(3.00×10−10m)2+0.08ke2(2.00×10−10m)2)×(11.0×10−10m)ke2d2 = (0.02004)×((8.988×109N.m2/C2)×(1.6×10−19C)2(1.0×10−10m)2)=4.6×10−10N
Conclusion:
The net force is 4.6×10-10N .
(b)
To calculate: The net force between the cytosine and guanine.
(b)
Answer to Problem 51P
The net force is 7.1×10−10N .
Explanation of Solution
Given:
The given helix-shaped DNA molecule
The net average charge on hydrogen and nitrogen atom is 0.2e , on carbon and oxygen atom is 0.40e and the separation distance between each molecule is 1.0×10-10m
Formula used:
The net force formula is given as,
F= kq1q2r2
Where, k is the proportionally constant, q1and q2 are testing charges and r is the separation distance between the testing charges.
The net force is calculated as,
FC.G= 2FO−H−2FO−N−FH−N+FN−N
Calculation:
Draw the structure explaining the bond between the cytosine and guanine.
Here, the bond which are responsible are OHN, NHN, and NHO.
For OHN bond:
For the O-H attraction, FO−H= k(0.4e)×(0.2e)(1.90×10−10m)2=0.08ke2(1.90×10−10m)2
For the H-N attraction FH−N= k(0.2e)×(0.2e)(2.00×10−10m)2=0.04ke2(2.00×10−10m)2
For NHN bond:
For the H-N attraction FH−N= k(0.2e)×(0.2e)(2.00×10−10m)2=0.04ke2(2.00×10−10m)2
For the N-N attraction, FN−N= k(0.2e)×(0.2e)(3.00×10−10m)2=0.04ke2(3.00×10−10m)2
For NHO bond:
For the O-N repulsion, FO−N= k(0.4e)×(0.2e)(2.90×10−10m)2=0.08ke2(2.90×10−10m)2
For the H-N attraction FH−N= k(0.2e)×(0.2e)(2.00×10−10m)2=0.04ke2(2.00×10−10m)2
The net force is given as,
FC.G= 2FO−H+FO−N+FH−N+FN−N = (20.08ke2(1.90×10−10m)2−20.08ke2(2.90×10−10m)2−0.08ke2(2.00×10−10m)2+0.08ke2(3.00×10−10m)2)×(11.0×10−10m)ke2d2 = (0.03085)×((8.988×109N.m2/C2)×(1.6×10−19C)2(1.0×10−10m)2)=7.1×10−10N
Conclusion:
The net force is 7.1×10−10N
(c)
To calculate: The total force for the DNA molecule.
(c)
Answer to Problem 51P
The net force is 6×10-5N .
Explanation of Solution
Given:
The number of pairs of molecules is 105 .
Formula used:
The net force is given as,
Fnet= 12(105)(FC.G+FA.T)
Calculation:
The total force is calculated by using the given value is,
Fnet= 12(105)(FC.G+FA.T) = (0.5)( 4.6×10-10N+7.1×10-10N) = 6×10−5N
Conclusion:
The net force is 6×10−5N .
Chapter 16 Solutions
Physics: Principles with Applications
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