Physics Ncert Solutions Class 12th

Physics Ncert Solutions Class 12th Class 12th

Contributor-Level 10

According to question, we can write

10 = $\frac{1}{2} a t^{2} . . . . . . . . . . . . . . . (1) a n d$

$10 + x = \frac{1}{2} a {(2 t)}^{2} . . . . . . . . . . . . . . . . (2)$

$\Rightarrow X = \frac{1}{2} A [{(2 t)}^{2} - t^{2}] = 3 (\frac{1}{2} a t^{2}) = 30 m$

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3 months ago

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

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Physics Ncert Solutions Class 12th Class 12th

Contributor-Level 10

Given, $λ = 8 m m, v = 8 m m, v = 3 * 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 * 1 0^{- 3}} = \frac{π}{4} * 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{j} v / m$

Now $\frac{E_{0}}{B_{0}} = C$

B0 = $\frac{60}{3 * 1 0^{8}} = \frac{20}{1 0^{8}} = 2 * 1 0^{- 7}$ $\vec{B} = 2 * 1 0^{- 7} [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{k} T$

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A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circuited.)

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Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

Contributor-Level 10

$P = \frac{E^{2}}{R} = \frac{{(N A \frac{d B}{d t})}^{2} * 4 A C}{ρ l}$

$p' = {(\frac{N A}{2} \frac{d B}{d t})}^{2} * 4 A C$

$p' = 2 P$

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3 months ago

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\hat{n}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $σ \hat{n}$ / $ε_{0}$ .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed lo

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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

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Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

Contributor-Level 10

2.16 Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body is $E_{2} .$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

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$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

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3 months ago

2.31 (a) Two large conducting spheres carrying charges $Q_{1}$ and $Q_{2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_{1} Q_{2}$ /4 $π ε_{0} r^{2}$ , where r is the distance between their centres?

(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if t

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(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

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2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elli

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2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

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4 months ago

13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

(a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_{1}^{2}{H)}$ =2.014102 u

m( ${}_{1}^{3}{H)}$ =3.016049 u

(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 = Boltzmann's constant, T = absolute temperature.)

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13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $* 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $* 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $* 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 * 10^{- 19})}^{2}}{4 * 10^{- 15}}$ $*$ 9 $* 10^{9}$ = 5.76 $* 10^{- 14}$ J =

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13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $* 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $* 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $* 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 * 10^{- 19})}^{2}}{4 * 10^{- 15}}$ $*$ 9 $* 10^{9}$ = 5.76 $* 10^{- 14}$ J = $\frac{5.76 * 10^{- 14}}{1.6 * 10^{- 19}}$ eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $* \frac{3}{2} k T$

Where

k = Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{K E}{3 k}$ = $\frac{5.76 * 10^{- 14}}{3 * 1.38 * 10^{- 23}}$ = 1.39 $* 10^{9}$ K

Hence the gas must be heated to 1.39 $* 10^{9}$ K to initiate the reaction.

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4 months ago

13.27 Consider the fission of ${}_{92}^{238}U$ 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 ${}_{58}^{140}{C e}$ and ${}_{44}^{99}{R u})$ . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${}_{92}^{238}U)$ =238.05079 u

m( ${}_{58}^{140}{C e}$ ) =139.90543 u

m( ${}_{44}^{99}{R u})$ = 98.90594 u

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13.27 In the fission of ${}_{92}^{238}U$ , $β^{-}$ 10 particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U + {}_{0}^{1}{n \to {}_{58}^{140}{C e +}} {}_{44}^{99}{R u + 10 {}_{- 1}^{0}e}$

It is given that:

Mass of a ${}_{92}^{238}U n u c l e u s, m_{1}$ = 238.05079 u

Mass of a ${}_{58}^{140}{C e} n u c l e u s, m_{2}$ =139.90543 u

Mass of a ${}_{44}^{99}{R u} n u c l e u s, m_{3}$ = 98.90594 u

Mass of a neutron ${}_{0}^{1}n$ , $m_{4}$ = 1.00865 u

Q value of the above equation,

Q = $[m^{'} ({}_{92}^{238}U) + m' ({}_{0}^{1}{n) - m^{'} (} {}_{58}^{140}{C e) - m^{'} ({}_{44}^{99}{R u) - 10 m_{e}}}] c^{2}$

Where

m' = represents the corresponding atomic masses of the nuclei.

$m^{'} ({}_{92}^{238}U)$ = $m_{1} - 92 m_{e}$

m'( ${}_{58}^{140}{C e)}$ = $m_{2} - 58 m_{e}$

m'( ${}_{44}^{99}{R u)}$ = $m_{3} - 44 m_{e}$

m'( ${}_{0}^{1}{n)} = m_{4}$

Substituting these values, we get

Q = $[m_{1} - 92 m_{e} + m_{4} - m_{2} + 58 m_{e} - m_{3} + 44 m_{e} - 10 m_{e}] c^{2}$

= $[m_{1} + m_{4} - m_{3} - m_{2}] c^{2}$

= $[238.05079 + 1.00865 - 98.90594 - 139.90543] c^{2}$ u

=0.24807 $c^{2}$ u

= 231.077 MeV

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13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $α -$ particle. Consider the following decay processes:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

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13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ &n

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13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 219.00948 - 4.00260) $c^{2}$ u

= 6.42 $* 10^{- 3} c^{2}$ u

= 6.42 $* 10^{- 3} * 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

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13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}^{41}{C a}$ and from the following data:

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

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13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $* 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.

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13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $* 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

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13.15 The Q value of a nuclear reaction A + b $\to$ C + d is defined by

Q = [ $m_{A} + m_{b}$ – $m_{C}$ – $m_{d}$ ] $c^{2}$

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) ${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

(ii) ${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic masses are given to be

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

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