Question:

The half-life of 3890​Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer:

Step 1: Calculate the amount of 3890Sr in 15mg.

15 mg = 0.015 g

Step 2: Calculate the number of moles of 3890Sr in 15mg.

0.015 g = 0.015/87.62 = 0.00017 moles of 3890Sr

Step 3: Calculate the number of atoms of 3890Sr in 15mg.

0.00017 moles x 6.022 x 10^23 atoms/mole = 1.02 x 10^21 atoms of 3890Sr

Step 4: Calculate the disintegration rate of 3890Sr in 15mg.

1.02 x 10^21 atoms/28 years = 3.6 x 10^19 atoms/year

Question:

The Q value of a nuclear reaction, A + b → C + d, is defined by Q=[mA​+mb​]−[mC​+md​] where, the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11​H+31​H→12​H+12​H (ii) 612​C+612​C→1020​Ne+24​He Atomic masses are given to be m(12​H)=2.014102u m(13​H)=3.016049u m(612​C)=12.000000u m(1020​Ne)=19.992439u

Answer:

(i) Q = [m(11H) + m(31H)] - [m(12H) + m(12H)] = [11.009305u + 93.018350u] - [2.014102u + 2.014102u] = 104.033657u - 4.028204u = 100.005453u

The reaction is exothermic as the Q-value is positive.

(ii) Q = [m(612C) + m(612C)] - [m(1020Ne) + m(24He)] = [12.000000u + 12.000000u] - [19.992439u + 24.000000u] = 24.000000u - 43.992439u = -19.992439u

The reaction is endothermic as the Q-value is negative.

Question:

Obtain the binding energy of the nuclei 2656​Fe and 83209​Bi in units of MeV from the following data: (a) m(2656​Fe)=55.934939u (b) m(83209​Bi)=208.980388u

Answer:

Step 1: Convert the given masses of 2656Fe and 83209Bi into kg by using the following formula:

m(2656Fe) = 55.934939u x 1.66054 x 10^-27 kg/u = 9.3118 x 10^-25 kg

m(83209Bi) = 208.980388u x 1.66054 x 10^-27 kg/u = 3.4737 x 10^-23 kg

Step 2: Calculate the binding energy of 2656Fe and 83209Bi using the following formula:

B.E(2656Fe) = m(2656Fe)c^2 - m(2656Fe)c^2 = 9.3118 x 10^-25 kg x (3 x 10^8 m/s)^2 - 9.3118 x 10^-25 kg x (931.5 MeV/c^2)

B.E(83209Bi) = m(83209Bi)c^2 - m(83209Bi)c^2 = 3.4737 x 10^-23 kg x (3 x 10^8 m/s)^2 - 3.4737 x 10^-23 kg x (931.5 MeV/c^2)

Step 3: Convert the binding energy of 2656Fe and 83209Bi into MeV by dividing the binding energy by 1.6022 x 10^-13 MeV/J.

B.E(2656Fe) = 8.834 x 10^-12 J / 1.6022 x 10^-13 MeV/J = 5.521 MeV

B.E(83209Bi) = 3.291 x 10^-10 J / 1.6022 x 10^-13 MeV/J = 205.3 MeV

Therefore, the binding energy of 2656Fe and 83209Bi in units of MeV is 5.521 MeV and 205.3 MeV respectively.

Question:

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive 614​C present with the stable carbon isotope 612​C. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of 614​C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 614​C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Answer:

1. The half-life of 614C is 5730 years.

2. The activity of the specimen from Mohenjodaro is 9 decays per minute per gram of carbon.

3. The age of the specimen can be calculated using the following equation: Age = (Half-life x ln(Activity/Original Activity))/ln2

4. Therefore, the approximate age of the Indus-Valley civilisation is (5730 x ln(9/15))/ln2 = 3650 years.

Question:

The nucleus 1123​Ne decays by β− emission. Write down the β - decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m(1023​Ne)=22.994466u m(1123​Na)=22.089770u.

Answer:

Beta- decay equation: 1123 Ne → 1123 Na + e- + ν

Maximum Kinetic Energy of Electron = [m(1123 Ne) - m(1123 Na) - m(e-)]c2

Maximum Kinetic Energy of Electron = [22.994466u - 22.089770u - 0.000548u]c2

Maximum Kinetic Energy of Electron = 0.904048u c2

Question:

(a) Two stable isotopes of lithium 36​Li and 37​Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. b) Boron has two stable isotopes, 510​B and 511​B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 510​B and 511​B B.

Answer:

a) The atomic mass of lithium is calculated by multiplying the abundance of each isotope by its mass and adding the two values together.

7.5% x 6.01512 u = 0.450864 u

92.5% x 7.01600 u = 6.54240 u

Total atomic mass of lithium = 7.013264 u

b) The abundance of each isotope is calculated by dividing its mass by the atomic mass of boron and multiplying by 100.

510B: 10.01294 u/10.811 u x 100 = 92.9%

511B: 11.00931 u/10.811 u x 100 = 101.1%

Question:

A source contains two phosphorous radio nuclides 1532​P(T1/2​=14.3d) and 1533​P(T1/2​=25.3d). Initially, 10% of the decays come from 1532​P. How long one must wait until 90% to do so?

Answer:

1. Calculate the decay rate of 1532P (decays per day) by using the formula ln(2)/T1/2 = 0.0493 decays/day.

2. Calculate the decay rate of 1533P (decays per day) by using the formula ln(2)/T1/2 = 0.0279 decays/day.

3. Calculate the initial amount of 1532P (in decays per day) by multiplying the decay rate by 10%, which is 0.00493 decays/day.

4. Calculate the initial amount of 1533P (in decays per day) by subtracting the initial amount of 1532P from the total decay rate, which is 0.0286 decays/day.

5. Calculate the time (in days) it takes for 1532P to reach 90% of the total decays by using the formula t = ln(0.1)/(0.0493) = 14.1 days.

6. Calculate the time (in days) it takes for 1533P to reach 90% of the total decays by using the formula t = ln(0.1)/(0.0279) = 25.2 days.

7. Add the time it takes for 1532P and 1533P to reach 90% of the total decays, which is 39.3 days.

Question:

Consider the fission of 92238​U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 58140​Ce and 4499​Ru. Calculate Q for this fission process. The relevant atomic and particle masses are: m(92238​U)=238.05079u m(58140​Ce)=139.90543u m(4499​U)=98.90594u.

Answer:

1. Calculate the mass of the reactants: m(92238U) = 238.05079u

2. Calculate the mass of the products: m(58140Ce) = 139.90543u, m(4499Ru) = 98.90594u

3. Calculate the mass defect: Δm = (m(reactants) - m(products)) = (238.05079u - (139.90543u + 98.90594u)) = -0.0445u

4. Calculate the binding energy: BE = Δm * c^2 = (-0.0445u * (3*10^8)^2) = -2.59 * 10^14 J

5. Calculate the Q value: Q = BE / (1.602 * 10^-19 J/eV) = (-2.59 * 10^14 J) / (1.602 * 10^-19 J/eV) = -1.62 * 10^33 eV

Question:

The fission properties of 94239​Pu are very similar to those of 92235​U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 94239​Pu undergo fission?

Answer:

Answer: Step 1: Calculate the number of atoms in 1 kg of 94239Pu.

1 kg of 94239Pu is equal to 6.022 x 10^26 atoms.

Step 2: Calculate the total energy released by all the atoms in 1 kg of 94239Pu.

The total energy released is equal to 6.022 x 10^26 atoms x 180 MeV = 1.084 x 10^29 MeV.

Question:

The radio nuclide 11C decays according to 611​C→511​B+e++ν: T1/2​ =20.3 min The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values: m(611​C)=11.011434u and m(611​B)=11.009305u Calculate Q and compare it with the maximum energy of the positron emitted.

Answer:

Step 1: Calculate the difference in mass between 611C and 511B, which is the difference in mass of the reactants and products.

m(611C) - m(511B) = 0.002130u

Step 2: Calculate Q, the energy released in the reaction, using the formula Q = Δm * c2, where c is the speed of light.

Q = 0.002130u * (3 x 108m/s)2 = 1.964 x 10-10 J

Step 3: Compare the energy released (Q) with the maximum energy of the positron emitted (0.960 MeV).

1.964 x 10-10 J = 0.960 MeV

The energy released (Q) is much smaller than the maximum energy of the positron emitted (0.960 MeV).

Question:

Write nuclear reaction equations for : (i) α - decay of 88226​Ra (ii) α - decay of 94242​Pu(iii) β− - decay of 1532​P (iv) β− - decay of 83210​Bi(v) β+ - decay of 611​C (vi) β+ - decay of 4397​Tc(vii) Electron capture of 54120​Xe

Answer:

(i) 88226Ra → 88224Rn + 4He

(ii) 94242Pu → 94239U + 4He

(iii) 1532P → 1533S + 0-1e

(iv) 83210Bi → 83211B + 0-1e

(v) 611C → 612N + 0+1e

(vi) 4397Tc → 4398Ru + 0+1e

(vii) 54120Xe + 0-1e → 54120Ba

Question:

From the relation R = R0​A1​/3 , where R0​ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Answer:

1. Rearrange the equation to isolate R on one side: R = (R0​A) / 3

2. Take the derivative of both sides with respect to A: dR/dA = (R0​) / 3

3. Since the derivative of a constant is zero, the right-hand side of the equation is equal to zero: dR/dA = 0

4. This means that the nuclear matter density is nearly constant, as it is independent of A.

Question:

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes: 88223​Ra→82209​Pb+614​C 88223​Ra→86219​Rn+24​He Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer:

1. Calculate the mass of each particle involved in the decays:

88223Ra = 223.0181 amu 82209Pb = 208.9811 amu 614C = 6.0151 amu 86219Rn = 218.0155 amu 24He = 4.0026 amu

2. Calculate the mass difference between the initial and final states (Δm):

For 88223Ra→82209Pb+614C: Δm = 223.0181 - (208.9811 + 6.0151) = 8.0229 amu

For 88223Ra→86219Rn+24He: Δm = 223.0181 - (218.0155 + 4.0026) = 1.00 amu

3. Calculate the Q-values for the decays:

For 88223Ra→82209Pb+614C: Q-value = Δm x c2 = 8.0229 x 931.5 MeV/amu = 7.507 MeV

For 88223Ra→86219Rn+24He: Q-value = Δm x c2 = 1.00 x 931.5 MeV/amu = 0.931 MeV

4. Determine that both are energetically allowed:

Since both Q-values are positive, both decays are energetically allowed.

Question:

Consider the DT reaction (deuteriumtritium fusion). 12​H+13​H→24​He+n (a) Calculate the energy released in MeV in this reaction from the data: m(12​H)=2.014102u m(13​H)=3.016049u (b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmans constant, T = absolute temperature.)

Answer:

(a) The energy released in this reaction can be calculated using the equation E=mc^2, where m is the difference between the masses of the reactants and products and c is the speed of light.

m(12H)=2.014102u m(13H)=3.016049u m(24He)=4.002603u

m(reactants)=5.010151u m(products)=4.002603u

E=(5.010151-4.002603)c^2 E=1.007548c^2 E=1.007548*(3x10^8)^2 E=8.6x10^13J

E=8.6x10^13J/1.602x10^-13MeV E=5.37x10^-13MeV

(b) The kinetic energy needed to overcome the coulomb repulsion between the two nuclei is equal to the average thermal kinetic energy available with the interacting particles. This can be calculated using the equation KE = 2(3kT/2), where k is Boltzmann’s constant and T is the absolute temperature.

KE = 2(3kT/2) KE = 3kT

T = KE/(3k) T = (5.37x10^-13MeV)/(3*1.38x10^-23J/K) T = 2.02x10^10K

Question:

For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K shell, is captured by the nucleus and a neutrino is emitted). e++ZA​X→Z−1A​Y+v Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice versa.

Answer:

Step 1: Understand the given statement.

Step 2: Recall that in β+ (positron) emission, a positron is emitted from the nucleus and the atomic number decreases by one.

Step 3: In electron capture, an electron from an inner orbit, such as the K shell, is captured by the nucleus and a neutrino is emitted. The atomic number increases by one.

Step 4: To show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice versa, we need to consider the energy difference between the initial and final states.

Step 5: In β+ emission, the energy difference between the initial and final states is equal to the energy of the emitted positron plus the energy of the neutrino.

Step 6: In electron capture, the energy difference between the initial and final states is equal to the energy of the captured electron plus the energy of the neutrino.

Step 7: Since the energy of the positron and the captured electron is the same, if β+ emission is energetically allowed, electron capture is necessarily allowed.

Step 8: However, if electron capture is energetically allowed, β+ emission may not be allowed.

Question:

The three stable isotopes of neon: 1020​Ne, 1021​Ne and 1022​Ne have respective abundances of 90.51%,0.27% and 9.22%. The atomic masses of the three isotopes are 19.99u,20.99u and 21.99u, respectively. Obtain the average atomic mass of neon.

Answer:

Step 1: Calculate the relative abundance of each isotope: 1020Ne: 90.51% 1021Ne: 0.27% 1022Ne: 9.22%

Step 2: Calculate the weighted average of the atomic masses: 1020Ne: 19.99u x 0.9051 = 18.05u 1021Ne: 20.99u x 0.0027 = 0.05u 1022Ne: 21.99u x 0.0922 = 2.02u

Step 3: Add up the weighted average of the atomic masses: 18.05u + 0.05u + 2.02u = 20.12u

Step 4: The average atomic mass of neon is 20.12u.

Question:

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200 MeV.

Answer:

Step 1: Calculate the total amount of thermal energy required to produce 200,000 MW of electric power.

Step 2: Multiply the total amount of thermal energy by 0.25 to calculate the amount of thermal energy that needs to be obtained from nuclear power plants.

Step 3: Calculate the amount of fissionable uranium needed per year to produce the required amount of thermal energy. This can be done by dividing the required thermal energy by the heat energy per fission of 235U (200 MeV).

Question:

Obtain approximately the ratio of the nuclear radii of the gold isotope 79197​Au and the silver isotope 47107​Ag .

Answer:

1. Find the atomic number of gold and silver. Gold: 79; Silver: 47

2. Find the mass number of gold and silver. Gold: 197; Silver: 107

3. Calculate the nuclear radius of gold and silver using the formula: R = r0A1/3 Where r0 = 1.2 fm and A is the mass number

Gold: R = 1.2 x 1971/3 = 8.2 fm Silver: R = 1.2 x 1071/3 = 6.6 fm

4. Calculate the ratio of the nuclear radii of gold and silver. Ratio = 8.2/6.6 = 1.24

Question:

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92235​U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92235​U and that this nuclide is consumed only by the fission process.

Answer:

1. Calculate the total energy generated by the reactor: 1000 MW x 5.00 y x 0.80 = 4,000,000 MW-y

2. Calculate the amount of 92235U consumed by the fission process: 4,000,000 MW-y / 3.2 x 10^13 MeV-y/mol = 1.25 x 10^7 mol

3. Calculate the amount of 92235U initially contained in the reactor: 1.25 x 10^7 mol x 238 g/mol = 297,500 g

Question:

Calculate and compare the energy released by (a) Fusion of 1.0 kg of hydrogen deep within Sun (b) The fission of 1.0 kg of 235U in a fission reactor

Answer:

a) The energy released by the fusion of 1.0 kg of hydrogen deep within the Sun is approximately 4.3 x 10^17 joules.

b) The energy released by the fission of 1.0 kg of 235U in a fission reactor is approximately 8.2 x 10^13 joules.

Therefore, the energy released by the fusion of 1.0 kg of hydrogen deep within the Sun is approximately 527 times greater than the energy released by the fission of 1.0 kg of 235U in a fission reactor.

Question:

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energy of the nuclei 1327​Al from the following data: m(1326​Al)=25.986895u m(1327​Al)=26.981541u

Answer:

1. Calculate the mass defect of 1327​Al: m(1327​Al) - m(1326​Al) = 0.994656u

2. Calculate the energy equivalent of the mass defect: E = mc^2 = 0.994656u x (931.5 MeV/u) = 925.2 MeV

3. Calculate the neutron separation energy of 1327​Al: S_n = 925.2 MeV

Question:

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 2963​Cu atoms (of mass 62.92960 u).

Answer:

Step 1: Determine the number of atoms in the coin.

Mass of the coin (3.0 g) / Mass of one Cu atom (62.92960 u) = 47.735 atoms

Step 2: Calculate the total number of neutrons and protons in the coin.

Number of neutrons and protons = Number of atoms x Number of neutrons and protons in one atom

= 47.735 x (29 + 63) = 4,096.755

Step 3: Calculate the nuclear energy required to separate all the neutrons and protons.

Nuclear energy = Total number of neutrons and protons x Energy required to separate one neutron and proton

= 4,096.755 x 1.00 x 10^-13 J = 4.096755 x 10^-13 J

Question:

Obtain the binding energy (in MeV) of a nitrogen nucleus (714​N). [Given: m (714​N) =14.00307 u.]

Answer:

Step 1: Convert the mass of the nitrogen nucleus from atomic mass units (u) to MeV.

1 u = 931.5 MeV

14.00307 u = 13041.8 MeV

Step 2: Calculate the binding energy of the nitrogen nucleus using the formula:

Binding energy = (mass of nucleus - mass of protons - mass of neutrons) x 931.5 MeV

Binding energy = (13041.8 MeV - 714 MeV - 714 MeV) x 931.5 MeV

Binding energy = 7113.3 MeV

Question:

Suppose, we think of fission of a 2656​Fe nucleus into two equal fragments of 1328​Al. Is the fission energetically possible? Argue by working out Q of the process. Given m (2656​Fe) = 55.93494 u and m (1328​Al) = 27.98191 u.

Answer:

Step 1: Calculate the total mass of the reactants. m(reactants) = m(2656Fe) + m(1328Al) = 55.93494 u + 27.98191 u = 83.91685 u

Step 2: Calculate the total mass of the products. m(products) = m(1328Al) + m(1328Al) = 27.98191 u + 27.98191 u = 55.96382 u

Step 3: Calculate the mass defect. mass defect = m(reactants) - m(products) = 83.91685 u - 55.96382 u = 27.95303 u

Step 4: Calculate the binding energy. Binding energy = mass defect x c^2 = 27.95303 u x (3 x 10^8 m/s)^2 = 2.45 x 10^17 J

Step 5: Calculate the Q value. Q = Binding energy of products - Binding energy of reactants = 2.45 x 10^17 J - 0 J = 2.45 x 10^17 J

Step 6: Conclusion. Since Q > 0, the fission of a 2656Fe nucleus into two equal fragments of 1328Al is energetically possible.

Question:

Find the Q-value and the kinetic energy of the emitted α- Particle in the α-decay of (a) 86220​Rn and (b) 88226​Ra. Given: m(88226​Ra)=226.02540u, m(86222​Rn)=222.01750u, m(84216​Po)=216.00189u

Answer:

a) Q-value = m(88226Ra) - m(86222Rn) - m(84216Po) = 226.02540u - 222.01750u - 216.00189u = 8.01101u

Kinetic energy of the emitted α-particle = Q-value = 8.01101u

b) Q-value = m(88226Ra) - m(86222Rn) - m(84216Po) = 226.02540u - 222.01750u - 216.00189u = 8.01101u

Kinetic energy of the emitted α-particle = Q-value = 8.01101u

Question:

In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 1224​Mg (23.98504u), 1225​Mg (24.98584u) and 1226​Mg (25.98259u). The natural abundance of 1224​Mg is 78.99% by mass. Calculate the abundances of other two isotopes.

Answer:

1. Calculate the total mass of the three isotopes: 1224Mg mass = 23.98504u 1225Mg mass = 24.98584u 1226Mg mass = 25.98259u

Total mass = 74.95447u

2. Calculate the abundance of 1224Mg: Abundance of 1224Mg = (23.98504u / 74.95447u) x 100 = 31.98%

3. Calculate the abundance of 1225Mg: Abundance of 1225Mg = (24.98584u / 74.95447u) x 100 = 33.31%

4. Calculate the abundance of 1226Mg: Abundance of 1226Mg = (25.98259u / 74.95447u) x 100 = 34.71%

Question:

Obtain the amount of 2760​Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 2760​Co is 5.3 years.

Answer:

Step 1: Calculate the decay constant (λ) of 2760​Co. λ = ln(2)/(half-life) λ = ln(2)/(5.3 years) λ = 0.1312year⁻¹

Step 2: Calculate the activity (A) of 8.0 mCi. A = 8.0 mCi

Step 3: Calculate the number of disintegrations (N) per second. N = A/λ N = 8.0 mCi / 0.1312year⁻¹ N = 61.1 x 10⁹ disintegrations/second

Step 4: Calculate the amount of 2760​Co in moles. N = N₀e⁻λt N₀ = N/e⁻λt N₀ = 61.1 x 10⁹ disintegrations/second / e⁻(0.1312year⁻¹)(5.3 years) N₀ = 2.80 x 10⁹ moles 2760​Co

Question:

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125% (b) 1% of its original value?

Answer:

a) Step 1: Calculate the number of half-lives it takes to reduce activity to 3.125%. 3.125% of the original value is equal to 1/32 of the original value. Therefore, it takes 5 half-lives to reduce the activity to 3.125%.

Step 2: Calculate the time it takes to reduce the activity to 3.125%. Since the half-life is T years, it will take 5T years to reduce the activity to 3.125%.

b) Step 1: Calculate the number of half-lives it takes to reduce activity to 1%. 1% of the original value is equal to 1/100 of the original value. Therefore, it takes 7 half-lives to reduce the activity to 1%.

Step 2: Calculate the time it takes to reduce the activity to 1%. Since the half-life is T years, it will take 7T years to reduce the activity to 1%.

Question:

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as : 12​H+12​→23​He+n+3.27MeV.

Answer:

Step 1: Calculate the energy released by the fusion reaction.

Energy released = 3.27 MeV

Step 2: Calculate the amount of energy required to keep the electric lamp glowing.

Energy required = 100 W x 3600 seconds = 360,000 J

Step 3: Calculate the amount of deuterium needed to provide the required energy.

Energy released per mole of deuterium = 3.27 MeV = 5.48 x 10^-11 J

Amount of deuterium needed = 360,000 J / 5.48 x 10^-11 J = 6.56 x 10^20 moles

Step 4: Calculate the mass of deuterium needed.

Mass of deuterium needed = 6.56 x 10^20 moles x 2.0 g/mole = 1.312 x 10^22 g

Step 5: Calculate the time the electric lamp can be kept glowing.

Time = Mass of deuterium needed / Energy released per mole of deuterium = 1.312 x 10^22 g / 5.48 x 10^-11 J = 2.39 x 10^32 seconds

Therefore, the electric lamp can be kept glowing for 2.39 x 10^32 seconds.

Question:

Calculate the height of the potential barrier for a head on collision of two deuterons.

Answer:

1. Determine the masses of the two deuterons.

2. Calculate the total kinetic energy of the two deuterons.

3. Determine the potential energy of the potential barrier between the deuterons.

4. Calculate the height of the potential barrier by subtracting the total kinetic energy from the potential energy.