The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the
geometric shape of these molecules, adenine bonds with thymine and cytosine bonds with guanine. The figure (Figure 1) shows the thymine-adenine bond. Each charge shown is ±e, and the H−N distance is 0.110 nm.
A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.
B) Is the net force attractive or repulsive?
C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.
D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.
A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.
B) Is the net force attractive or repulsive?
C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.
D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.
1 answer:
Answer:
Explanation:
An image of the bond resulting from the search is attached.
Consider the force directed towards thymine as negative.
For the O-H-N combination:
The resulting force from this combination is:
In the case of the N-H-N combination:
The total force acting from this combination is:
The force that thymine applies on adenine is:
When rounded to three significant figures, the net force is
b)
The negative value indicates that the force is attractive, as it is aimed towards thymi.
c)
The force acting on the electron due to the proton is:
Since the electron and proton carry opposite charges, the force on the electron points towards the proton.
d)
The ratio of the above forces is:
Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.
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Answer:
Refer to the explanation
Explanation:
Solution:-
- For this problem, we will assume the grating has a defined line density ( N ) representing the lines per mm.
- The angle formed by each fringe on the screen is given by ( θ ).
- The order of bright/dark spots is characterized by an integer ( n )
- The incident light's wavelength is ( λ )
- Using the diffraction grating relationship from the Young's Experiment is as follows:
- The provided formula corresponds to constructive interference.
- We will examine how increasing the distance between the screen and the grating affects the pattern. Let ( L ) be the distance from the grating center to the screen center. The distance ( yn ) indicates the vertical spacing between each fringe on the screen.
- For small angles ( θ ), we use the approximation of sin ( θ ) ≈ tan ( θ ). Thus,
sin ( θ ) ≈ tan ( θ ) = [ yn / L ]
- Replacing this approximation in the diffraction grating relation results in:
- To double the distance from the screen to the grating, we use the relation with ( 2L ):
yn ∝ L
Result: The separation between each order of bright and dark fringe doubles. The interference pattern will spread further! Consequently, there will be fewer bright spots visible on the screen since a larger surface area is required to accommodate the entire pattern. The increased distance also diminishes the intensity contrast between bright and dark fringes due to the extended path traveled by light rays. Intensity is inversely related to the square of the travel distance.
- If the line density of the grating ( N ) were doubled, it follows that:
yn ∝ N
Result: The spacing between bright and dark fringes would also double. The interference pattern expands further, leading to the necessity of a bigger screen to show the entire pattern.
Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
Given
Process 1 ---> 2
Pressure at point (1) P1 = 10 bar = P2
Volume at point (1) V1 = 1 m^3
Volume at point (2) V2 =4 m^3
For Process 2 ---> 3, where V = constant
Pressure at point (3) P3 = 10 bar
Volume at point (3) V3 = 4 m^3
Process 3 ---> 1 defined as PV = constant.
Required
Sketch the processes on the PV coordinates To calculate the work in kJ
Work is calculated by W=
a=V2
b=V1
x=Pdx=dV
For Process 1 ---> 2 where P3 = P4 = 5 barW=
a=V3
b=V2x=4dx=dV
substituting the values here into the integral givesW=400 kJFor Process 2 ---> 3
As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ
For Process 3 ---> 1 By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1
W=a=V1
x=1V^-1dx=dVsubstituting values into this integral results in
W=| ln V | limits a and b= -160.944 KJ