Calculate the binding energy per nucleon of ₂₆Fe⁵⁶. Atomic mass of ₂₆Fe⁵⁶ is 55.9349 u and that of ₁H¹ is 1.00783u. Mass of ₀n¹ = 1.00867u and 1u = 931 MeV.

Solution
Given,
Atomic mass of ₂₆Fe⁵⁶ (M) = 55.9349 u.
Here, Z = 26 & A = 56
Mass of ₁H¹ (m_p) = 1.00783 u.
Mass of ₀n¹ (m_n) = 1.00867 u
1u = 931 MeV
We know that,
B.E. = [Zm_p + (A – Z) m_n – M]. 931 MeV
= [latex]
[26\times 1.00783 + (56 – 26)\times 1.00867 – 55.9349]\times 931 MeV
[/latex]
                  = 492.294 MeV
Now, B.E. per nucleon
[latex]
(\bar{B.E.}) = \frac{B.E.}{A} = \frac{492.294}{56}
[/latex]
= 8.791 Mev/nucleon.

Calculate the binding energy per nucleon of calcium nucleus (₂₀Ca⁴⁰).
Given: mass of ₂₀Ca⁴⁰ = 39.962589 u
Mass of neutron, m_n = 1.008665u
Mass of proton, m_p = 1.007825u
1u = 931 MeV.

Solution
Given,
Atomic mass of ₂₀Ca⁴⁰ (M) = 39.962589 u.
Here, Z = 20 & A = 40
Mass of ₁H¹ (m_p) = 1.007825 u.
Mass of ₀n¹ (m_n) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zm_p + (A – Z) m_n – M]. 931 MeV
= [latex][20\times 1.007825 + (40 – 20)\times 1.008665 – 39.962589]\times 931 MeV [/latex]
= 341.873 MeV
Now, B.E. per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{341.873}{40}[/latex] = 8.547 Mev/nucleon.

A city requires 10⁷ watts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency 20%. Using ₉₂U²³⁵ as the fuel source, calculate the amount of fuel required per day. (Energy released per fission ₉₂U²³⁵ = 200 MeV).

Solution
Given,
Electrical power required by city (P_out) = 10⁷ Watt
Efficiency [latex](\eta)[/latex] = 20%
Amount of fuel required per day (m) = ?
Energy released per fission (E) = 200 MeV = 200 x 1.6 x 10^-13
= 3.2 x 10^-11 J
We have, efficiency
[latex]
(\eta) = \frac{P_{out}}{P_{in}}\times 100%
[/latex]

Or, 20 % = [latex]\frac{P_{out}}{P_{in}}\times 100%[/latex]

Or, 0.2 = [latex]\frac{10^7}{P_{in}}[/latex]

Or, P_in = [latex]\frac{10^7}{0.2} = 5\times 10^7[/latex] Watt

Now, Time (t) = 1 day = 86400 Sec
So, Power = [latex]\frac{Total\ Energy\ (E_T)}{Time\ taken\ (t)}[/latex]

Or, E_T = P_in . t = 5 x 10⁷ x 86400
                  = 4.320 x 10¹² J
Now, No. of atom released per fission (N) = [latex]\frac{E_T}{E} = \frac{4.320\times 10^{12}}{3.2\times 10^{-11}}[/latex]
= [latex]1.350\times 10^{23}[/latex]
Using Avogadro’s Hypothesis,
∵ 6.023 x 1023 atom of U – 235 is contained in 235 g of it.
∴ 1 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}[/latex] g of it.
∴ 1.35 x 1023 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}\times 1.35\times 10^{23}[/latex] g of it.
            = 0.0527 Kg of it.

A nucleus of ₉₂U²³⁸ disintegrates according to
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴
Calculate:
(i) the total energy released in the disintegration process.
(ii) the K.E. of the [latex]\alpha[/latex] – particles, the nucleus at rest before disintegration. (Mass of ₉₂U²³⁸ = [latex]3.859\times 10^{-25}[/latex] kg, mass of ₉₀Th²³⁴ = [latex]3.787\times 10^{-25}[/latex] kg, mass of ₂He⁴ = [latex]6.648\times 10^{-27}[/latex] kg).

Solution
Given,
₉₂U²³⁵ disintegrates according to ₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴
We have,
Mass of ₉₂U²³⁸ = 3.859 x 10^-25 Kg
Mass of ₉₀Th²³⁴ = 3.787 x 10^-25 Kg
Mass of ₂He⁴ = 6.648 x 10^-27 Kg
Total energy released in the disintegration process (E)
= [latex][\sum_{reactant} – \sum_{product}]\times c^2[/latex]
= [3.859 x 10^-25 – (3.787 x 10^-25 + 6.648 x 10^-27)] x c²
= 4.968 x 10^-11 J
= 310.5 MeV
Now, K.E. of [latex]\alpha[/latex] – particles = [latex]\frac{M_{Th}}{M_{Th} + M_{He}}\times E[/latex]

                  = [latex]\frac{3.787\times 10^{-25}}{(3.787\times 10^{-25}) + (6.648\times 10^{-27})}\times 310.5[/latex]

            = 305.143 MeV.

The mass of ₁₇Cl³⁵ is 34.9800 amu. Calculate its binding energy and binding energy per nucleon. Mass of one proton = 1.007825 amu and mass of one neutron = 1.008665 amu.

Solution
Given,
Atomic mass of ₁₇Cl³⁵ (M) = 34.9800 u
Here, Z = 17 & A = 35
Mass of proton (m_p) = 1.007825 u.
Mass of neutron (m_n) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zm_p + (A – Z) m_n – M]. 931 MeV
                  = [latex][17\times 1.007825 + (35 – 17)\times 1.008665 – 34.98]\times 931 MeV [/latex]
                  = 287.674 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{287.674}{35}[/latex] = 8.219 Mev/nucleon.

The energy liberated in the fission of single uranium – 235 atom is [latex]3.2\times 10^{-11} J[/latex]. Calculate the power production corresponding to the fission of 1 gm of uranium per day. Assume Avogadro constant as [latex]6\times 10^{23} mol^{-1}[/latex].

Solution
Given,
Energy liberated in the fission of single uranium atom (E) = 3.2 x 10^-11 J
Power production (P) =?
Mass of uranium (m) = 1 gm = 1 x 10^-3 kg
Time (t) = 1 day = 86400 s
We know that,
According to Avogadro’s hypothesis,
[latex]\because[/latex] 235 gm of U – 235 contains [latex]6.023\times 10^{23}[/latex] atoms.
[latex]\therefore[/latex] 1 gm of U – 235 contains [latex]\frac{6.023\times 10^{23}\ atoms}{235}[/latex].
Or, N = 2.563 x 10²¹ atoms.
Total energy liberated (T.E.) = N.E
      = (2.563 x 10²¹) x (3.2 x 10^-11)
      = 8.202 x 10¹⁰ J
Power production (P) = [latex]\frac{T.E.}{t}[/latex]
      = [latex]\frac{8.202\times 10^{10}}{86400}[/latex]
            = 9.493 x 10⁵ Watt.

What will be the amount of energy released in the fusion of three alpha particles into a C¹² nucleus if mass of He⁴ and C¹² nuclei are respectively 4.00263 amu and 12 amu.

Solution
Given,
Amount of energy released (E) =?
Mass of He⁴ = 4.00263 u
Mass of C¹² = 12 u
We have, the reaction involved in the fusion is written as:
3₂He⁴ → ₆C¹² + Q
So, Q – energy = [latex][\sum_{reactant} – \sum_{product}]\times 931 MeV[/latex]
      = [(3 x 4.00263) – 12] x 931 MeV
     = 7.346 MeV.

The mass of the nucleus of the isotope Lithium ₃Li⁷ is 7.014351 u. Find its binding energy and binding energy per nucleon. (Given mass of proton = 1.00727 u, mass of neutron = 1.008665 u).

Solution
Given,
Atomic mass of ₃Li⁷ (M) = 7.014351 u
Here, Z = 3 & A = 7
Mass of proton (m_p) = 1.00727 u
Mass of neutron (m_n) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zm_p + (A – Z) m_n – M]. 931 MeV
= [latex][3\times 1.00727 + (7 – 3)\times 1.008665 – 7.014351]\times 931 MeV[/latex]
                  = 39.213 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{39.213}{7}[/latex] = 5.602 Mev/nucleon.

₂₈Ni⁶² may be described as the most strongly bound nucleus because it has the highest B.E. per nucleon. Its neutral atomic mass is 61.928349 amu. Find its mass defect, its total binding energy and binding energy per nucleon.
Given, mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
1 amu = 931.5 MeV.

Solution
Given,
Atomic mass of ₂₈Ni⁶² (M) = 61.928349 u
Here, Z = 28 & A = 62
Mass of proton (m_p) = 1.007825 u.
Mass of neutron (m_n) = 1.008665 u
1u = 931.5 MeV
We know that,
Mass defect [latex](\Delta m)[/latex] = [Zm_p + (A – Z) m_n – M]
= [latex][28\times 1.007825 + (62 – 28)\times 1.008665 – 61.928349][/latex]
      = 0.585361 amu
B.E. = [latex][\Delta m][/latex]. 931 MeV
                  = [latex][0.585361]\times 931.5\ MeV[/latex]
                  = 545.264 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{545.264}{62}[/latex] = 8.795 Mev/nucleon.

The energy released per fission of one U235 atom is 200 MeV. Calculate the energy released in Kwh, when one gram of uranium undergoes fission.

Solution

Given,

Energy liberated per fission (E) = 200 MeV

= 200 x (1.6 x 10-13) J

= 3.2 x 10-11 J

Energy released in Kwh =?

Mass of uranium (m) = 1 gm = 1 x 10-3 kg

Time (t) = 1 hour = 3600 s

We know that,

According to Avogadro’s hypothesis,

∵ 235 gm of U – 235 contains 6.023 × 1023 atoms.

∴ 1 gm of U – 235 contains [latex]\dfrac{6.023 \times 10^{23}}{235}[/latex] atoms.

N = 2.563 x 1021 atoms.

Total energy liberated (T.E.) = N.E

[latex]= (2.563 \times 10^{21}) \times (3.2 \times 10^{-11})[/latex]

[latex]= 8.202 \times 10^{10}\ J[/latex]

Power production [latex](P) = \dfrac{T.E.}{t}[/latex]

[latex]= \dfrac{8.202 \times 10^{10}}{3600}[/latex]

[latex]= 2.278 \times 10^7\ Watt[/latex]

[latex]= \dfrac{2.278 \times 10^7}{1000}\ KW[/latex]

[latex]= 2.278 \times 10^4\ KW.[/latex]

Calculate the binding energy per nucleon for a helium nucleus. Given that mass of helium nucleus = 4.001509 amu, mass of proton = 1.007277 amu and mass of neutron = 1.008666 amu.

Solution

Given,

Atomic mass of [latex]{}_2He^4[/latex] (M) = 4.001509 u

Here, Z = 2 & A = 4

Mass of proton [latex](m_p)[/latex] = 1.007277 u

Mass of neutron [latex](m_n)[/latex] = 1.008666 u

1u = 931 MeV

We know that, [latex]B.E. = [Zm_p + (A-Z)m_n – M] \times 931\ MeV[/latex]

[latex]= [2 \times 1.007277 + (4-2) \times 1.008666 – 4.001509] \times 931\ MeV[/latex]

[latex]= 28.281\ MeV[/latex]

Now, Binding Energy per nucleon [latex]\bar{B.E.} = \dfrac{B.E.}{A} = \dfrac{28.281}{4}[/latex]

[latex]= 7.07\ MeV/nucleon.[/latex]

The most common isotope of uranium 92U238, has atomic mass 238.050783 u. Calculate the a) mass defect, b) binding energy, c) Binding energy per nucleon. (Mass of proton = 1.007825 u, mass of neutron = 1.008665 u).

Solution

Given,

Atomic mass of [latex]{}_{92}U^{238}[/latex] (M) = 238.050783 u

Here, Z = 92 & A = 238

Mass of proton [latex](m_p)[/latex] = 1.007825 u

Mass of neutron [latex](m_n)[/latex] = 1.008665 u

1u = 931 MeV

We know that,

Mass defect [latex](\Delta m) = [Zm_p + (A-Z)m_n – M][/latex]

[latex]= [92 \times 1.007825 + (238-92) \times 1.008665 – 238.050783][/latex]

[latex]= 1.934207\ amu[/latex]

[latex]B.E. = [\Delta m] \times 931\ MeV[/latex]

[latex]= 1.934207 \times 931\ MeV[/latex]

[latex]= 1800.747\ MeV[/latex]

Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \dfrac{B.E.}{A} = \dfrac{1800.747}{238}[/latex]

[latex]= 7.566\ MeV/nucleon.[/latex]

Estimate the binding energy per nucleon of 3Li7. Mass of 3Li7, a proton & a neutron are respectively 7.01435 amu, 1.00728 amu and 1.00867 amu.

Solution

Given,

Atomic mass of [latex]{}_3Li^7[/latex] (M) = 7.01435 u

Here, Z = 3 & A = 7

Mass of proton [latex](m_p)[/latex] = 1.00728 u

Mass of neutron [latex](m_n)[/latex] = 1.00867 u

1u = 931 MeV

We know that, [latex]B.E. = [Zm_p + (A-Z)m_n – M] \times 931\ MeV[/latex]

[latex]= [3 \times 1.00728 + (7-3) \times 1.00867 – 7.01435] \times 931\ MeV[/latex]

[latex]= 39.26\ MeV[/latex]

Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \dfrac{B.E.}{A} = \dfrac{39.26}{7}[/latex]

[latex]= 5.609\ MeV/nucleon.[/latex]

Assuming that about 200 MeV energy is released per fission of [latex]{}_{92}U^{235}[/latex] nuclei. What would be the mass of [latex]U^{235}[/latex] consumed per day in the fission of power 1 MW approximately.

Solution

Given,

Energy liberated per fission (E) = 200 MeV

= [latex]200 \times (1.6 \times 10^{-13})[/latex]

= [latex]3.2 \times 10^{-11}[/latex] J

Mass of [latex]U^{235}[/latex] (m) =?

Time (t) = 1 day = 86400 s

Power (P) = 1 MW

= [latex]1 \times 10^6[/latex] W

We know that,

Power production [latex](P) = \dfrac{T.E.}{t}[/latex]

[latex]T.E. = P.t[/latex]

[latex]= 10^6 \times 86400[/latex]

[latex]= 8.640 \times 10^{10}\ J[/latex]

Total energy liberated (T.E.) = N.E

[latex]N = \dfrac{T.E.}{E}[/latex]

[latex]= \dfrac{8.64 \times 10^{10}}{3.2 \times 10^{-11}}[/latex]

[latex]= 2.7 \times 10^{21}\ J[/latex]

According to Avogadro’s hypothesis,

∵ [latex]6.023 × 10^{23}[/latex] atoms of U – 235 is contained in 235 gm of it.

∴ [latex]2.7 × 10^{21}[/latex] atoms of U – 235 is contained in [latex]\dfrac{235}{6.023 \times 10^{23}} \times (2.7 \times 10^{21})[/latex]

m = 1.053 gm.

A nucleus of uranium – 238 disintegrates according to the reaction [latex]{}_{92}U^{238} → {}_{90}Th^{234} + {}_2He^4[/latex]. Calculate: i) the total energy released in the disintegration ii) the K.E. of the alpha particle, the nucleus being at rest before disintegration. (Mass of [latex]{}_{92}U^{238} = 238.12492\ u[/latex], mass of [latex]{}_{90}Th^{234} = 234.1165\ u[/latex], mass of [latex]{}_2He^4[/latex] = 4.00387 u, 1u = 931 MeV).

Solution

Given,

[latex]{}_{92}U^{235}[/latex] disintegrates according to [latex]{}_{92}U^{238} → {}_{90}Th^{234} + {}_2He^4[/latex]

We have,

Mass of [latex]{}_{92}U^{238}[/latex] = 238.12492 u

Mass of [latex]{}_{90}Th^{234}[/latex] = 234.1165 u

Mass of [latex]{}_2He^4[/latex] = 4.00384 u

Total energy released in the disintegration process (E)

[latex]= [\sum reactant – \sum product] \times 931\ MeV[/latex]

[latex]= [238.12492 – (234.1165 + 4.00384)] \times 931\ MeV[/latex]

[latex]= 4.264\ MeV[/latex]

Now, K.E. of [latex]\alpha[/latex] – particles [latex]= \dfrac{M_{Th}}{M_{Th} + M_{He}} \times E[/latex]

[latex]= \dfrac{234.1165}{(234.1165+4.00384)} \times 4.264[/latex]

[latex]= 4.192\ MeV.[/latex]

The energy liberated in the fission of a single uranium – 235 atom is [latex]3.2 × 10^{−11}[/latex] J. Calculate the power production corresponding to the fission of 1 kg of uranium per day. (Avogadro constant = [latex]6.0 × 10^{23} mole^{−1}[/latex]).

Solution

Given,

Energy liberated per fission (E) = [latex]3.2 \times 10^{-11}[/latex] J

Power production (P) =?

Mass of uranium (m) = 1 kg = 1000 gm

Time (t) = 1 day = 86400 s

Avogadro’s constant [latex](N_A) = 6.0 \times 10^{23}\ mole^{-1}[/latex]

We know that,

According to Avogadro’s hypothesis,

∵ 235 gm of U – 235 contains [latex]6.0 \times 10^{23}[/latex] atoms.

∴ 1000 gm of U – 235 contains [latex]\dfrac{6.0 \times 10^{23}}{235} \times 1000[/latex] atoms

N = [latex]2.553 \times 10^{24}[/latex] atoms.

Total energy liberated (T.E.) = N.E

[latex]= (2.553 \times 10^{24}) \times (3.2 \times 10^{-11})[/latex]

[latex]= 8.17 \times 10^{13}\ J[/latex]

Power production [latex](P) = \dfrac{T.E.}{t}[/latex]

[latex]= \dfrac{8.17 \times 10^{13}}{86400}[/latex]

[latex]= 9.456 \times 10^8\ Watt.[/latex]

Calculate the Q – value of the reaction & mention the type of reaction (endo – thermic or exothermic). [latex]{}_2He^4[/latex] = 4.00377 amu, [latex]{}_8O^{17}[/latex] = 17.00450 amu, [latex]{}_7N^{14}[/latex] = 14.00783 amu, [latex]{}_1H^1[/latex] = 1.00814 amu.

Solution

Given,

Q – value of the reaction (Q) =?

Mass of [latex]{}_2He^4[/latex] = 4.00377 u

Mass of [latex]{}_8O^{17}[/latex] = 17.00450 u

Mass of [latex]{}_7N^{14}[/latex] = 14.00783 u

Mass of [latex]{}_1H^1[/latex] = 1.00814 u

We have, the reaction involved is written as:

[latex]{}_2He^4 + {}_7N^{14} → {}_8O^{17} + {}_1H^1[/latex] + Q

So, Q – value [latex]= [\sum reactant – \sum product] \times 931\ MeV[/latex]

[latex]= [(4.00377+14.00783) – (17.00450+1.00814)] \times 931\ MeV[/latex]

[latex]= -0.96824\ MeV[/latex]

Since, Q – value of the reaction is negative, the reaction is endothermic.

Calculate the speed of particle of the mass of it is equal to 5 times its rest value.

Solution

Given,

Speed of particle (v) =?

Mass of particle (m) = 5 times rest mass [latex](m_o)[/latex] i.e.

m = [latex]5m_o[/latex]

We know that,

From relativistic relation of mass,

[latex]m = \dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]

[latex]5m_o = \dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]

[latex]\sqrt{1-\dfrac{v^2}{c^2}} = \dfrac{1}{5}[/latex]

Squaring both sides,

[latex]1 – \dfrac{v^2}{c^2} = \dfrac{1}{25}[/latex]

[latex]\dfrac{v^2}{c^2} = 1 – \dfrac{1}{25}[/latex]

[latex]v^2 = \dfrac{24}{25} \times (3 \times 10^8)^2[/latex]

[latex]v^2 = 8.64 \times 10^{16}[/latex]

[latex]v = \sqrt{8.64 \times 10^{16}}[/latex]

[latex]= 2.939 \times 10^8\ ms^{-1}.[/latex]