Δ G 0 value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained. Concept introduction: P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700. In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level. An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

The standard cost of Product B manufactured by Oriole Company includes 3.5 units of direct materials at $5.40 per unit. During June, 27,300 units of direct materials are purchased at a cost of $5.15 per unit, and 27,300 units of direct materials are used to produce 7,600 units of Product B. (a) Compute the total materials variance and the price and quantity variances. Total materials variance Materials price variance Materials quantity variance $ (b) Compute the total materials variance and the price and quantity variances, assuming the purchase price is $6.35 and the quantity purchased and used is 26,300 units. Total materials variance Materials price variance Materials quantity variance $

The pyruvate dehydrogenase complex catalyzes the oxidative decarboxylation of pyruvate to form acetyl CoA. E₁, E2, and E3 are abbreviations for the enzymes of the complex. Classify the enzyme names, prosthetic groups, and reactions as E1, E2, or E3. E₁ E2 Answer Bank E3 transfer of electrons to FAD and then to NAD+ transfer of acetyl group to coenzyme A formation of hydroxyethyl-TPP hydroxyethyl group transferred to lipoamide thiamine pyrophosphate (TPP) FAD lipoamide dihydrolipoyl transacetylase pyruvate dehydrogenase dihydrolipoyl dehydrogenase

Patients with pyruvate dehydrogenase deficiency show high levels of lactic acid in the blood. However, in some cases, treatment with dichloroacetate (DCA), which inhibits the kinase associated with the pyruvate dehydrogenase complex, lowers lactic acid levels. How does DCA act to stimulate pyruvate dehydrogenase activity? DCA activates pyruvate dehydrogenase kinase. DCA increases phosphorylation levels of pyruvate dehydrogenase. DCA inhibits pyruvate dehydrogenase kinase. ODCA activates pyruvate dehydrogenase phosphatase. What does this suggest about pyruvate dehydrogenase activity in patients who respond to DCA? The pyruvate dehydrogenase complex is active only when phosphorylated by the kinase. The pyruvate dehydrogenase complex is active only in the presence of the kinase. The pyruvate dehydrogenase complex is completely inactive. The pyruvate dehydrogenase complex displays some residual activity.

Answer 2

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 6

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 7

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 12

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 13

Ch. 15 - Prob. 1P Ch. 15 - Prob. 2P Ch. 15 - Prob. 3P Ch. 15 - Prob. 4P Ch. 15 - The flux of solar energy reaching the Earth's...Ch. 15 - The substance dichlorophenyldimethylurea (DCMU) is...Ch. 15 - Prob. 7P Ch. 15 - Prob. 8P Ch. 15 - Prob. 9P Ch. 15 - Prob. 10P

Videos

Answer to Problem 1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 15 Solutions