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Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The X-Y plane be taken as the boundary between two transparent media M₁ and M₂. M₁ in $z \geq 0$ has a refractive index of $\sqrt[]{2}$ and M₂ with Z < 0 has refractive index of $\sqrt[]{3} .$ A ray of light travelling in M₁ along the direction given by the vector $\vec{P} = 4 \sqrt[]{3} \hat{i} - 3 \sqrt[]{3} \hat{j} - 5 \hat{k},$ is incident on the plane of separation. The value of difference between the angle of incident in M₁ and the angle of refraction in M₂ will be_______ degree.

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Vishal Baghel

Contributor-Level 10

For angle of incidence 'i' :-

cos I = $\frac{5}{\sqrt[]{48 + 27 + 25}} = \frac{5}{\sqrt[]{100}}$

i = 60°

Using snell's law :-

$μ_{1} s i n i = μ_{2} s i n r$

sin r = $\frac{\sqrt[]{2}}{\sqrt[]{3}} s i n 60 ° = \sqrt[]{\frac{2}{3}} * \frac{\sqrt[]{3}}{2} = \frac{1}{\sqrt[]{2}}$

$∴ r = 45 °$

So, difference, I – r = 60° - 452 = 15°

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3 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Find the binding energy per neucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = (A) 1.00783 U$ , mass of neutron $m_{n} = (A) 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ . (take $1 U = 931 M e V$ )

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alok kumar singh

Contributor-Level 10

Sol. $\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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3 months ago

Nuclei Class 12th

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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Payal Gupta

Contributor-Level 10

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 =

...more

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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3 months ago

Nuclei Class 12th

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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Payal Gupta

Contributor-Level 10

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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3 months ago

Nuclei Class 12th

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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Payal Gupta

Contributor-Level 10

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

...more

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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Nuclei Class 12th

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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Payal Gupta

Contributor-Level 10

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.

...more

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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Nuclei Class 12th

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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Payal Gupta

Contributor-Level 10

13.15 The given nuclear reaction is

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

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

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

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

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 * 2.014102] c^{2}$

= (-4.33 $* 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

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

Atomic mass

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

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

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 * 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $* 10^{- 3} c^{2}$ u

=4.958 $* 10^{- 3} * 931.5$

=4.6183 MeV

...more

13.15 The given nuclear reaction is

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

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

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

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

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 * 2.014102] c^{2}$

= (-4.33 $* 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

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

Atomic mass

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

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

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 * 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $* 10^{- 3} c^{2}$ u

=4.958 $* 10^{- 3} * 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

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Nuclei Class 12th

13.13 The radionuclide ${}_{6}^{11}C$ decays according to ${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v :}} T_{1 / 2}$ =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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Payal Gupta

Contributor-Level 10

13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m' = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

= $[11.011434 - 11.009305 - 2 * 0.000548] c^{2}$ u

=1.033

...more

13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m' = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

= $[11.011434 - 11.009305 - 2 * 0.000548] c^{2}$ u

=1.033 $* 10^{- 3} c^{2}$ u

But 1 u = 931.5 MeV/ $c^{2}$

ΔQ = 0.962 MeV

The value of $Δ Q$ is almost comparable to the maximum energy of the emitted positron.

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Nuclei Class 12th

13.12 Find the Q-value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of (a) ${}_{88}^{226}{R a}$ and (b) ${}_{86}^{220}{R n}$ . Given m ( ${}_{88}^{226}{R a})$ = 226.02540 u, m ( ${}_{86}^{222}{R n})$ = 222.01750 u, m ( ${}_{86}^{222}{R n})$ = 220.01137 u, m ( ${}_{84}^{216}{P o)}$ = 216.00189 u.

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Payal Gupta

Contributor-Level 10

13.12 $α -$ particle decay of ${}_{88}^{226}{R a}$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $* c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{88}^{226}{R a})$ = 226.02540 u

m ( ${}_{86}^{222}{R n})$ = 222.01750 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^{2}$

= 5.297 $* 10^{- 3} {u c}^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Hence Q = 4.934 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $* Q$ = $\frac{222}{226}$ $*$ 4.934= 4.85 MeV

$α -$ particle decay of

...more

13.12 $α -$ particle decay of ${}_{88}^{226}{R a}$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $* c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{88}^{226}{R a})$ = 226.02540 u

m ( ${}_{86}^{222}{R n})$ = 222.01750 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^{2}$

= 5.297 $* 10^{- 3} {u c}^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Hence Q = 4.934 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $* Q$ = $\frac{222}{226}$ $*$ 4.934= 4.85 MeV

$α -$ particle decay of ${}_{86}^{220}{R n}$ emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.

${}_{86}^{220}{R n}$ $\to$ ${}_{84}^{216}{P o}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $* c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{86}^{222}{R n})$ = 220.01137 u

m ( ${}_{84}^{216}{P o)}$ = 216.00189 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(220.01137) – (216.00189 + 4.002603)] $c^{2}$

= 6.877 $* 10^{- 3} {u c}^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Hence Q = 6.406 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $* Q$ = $\frac{216}{222}$ $*$ 6.406= 6.23 MeV

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New answer posted

3 months ago

Nuclei Class 12th

13.6 Write nuclear reaction equations for

(i) $α$ -decay of ${}_{88}^{226}{R a}$

(ii) $α$ -decay of ${}_{94}^{242}{P u}$

(iii) $β$ –-decay of ${}_{15}^{32}P$

(iv) $β$ –-decay of ${}_{83}^{210}{B i}$

(v) $β$ +decay of ${}_{6}^{11}C$

(vi) $β$ –-decay of ${}_{43}^{97}{T c}$

(vii) Electron capture of ${}_{54}^{120}{X e}$

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Payal Gupta

Contributor-Level 10

13.6 In $α$ - decay, there is a loss of 2 protons and 4 neutrons. In every $β +$ decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every $β -$ decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

${}_{94}^{242}{P u}$ $\to$ ${}_{92}^{238}U$ + ${}_{2}^{4}{H e}$

${}_{15}^{32}P$ $\to$ ${}_{16}^{32}S$ + $e^{-}$ + $\overset{?}{ν}$

${}_{83}^{210}{B i}$ $\to$ ${}_{84}^{210}{P o}$ + $e^{-}$ + $\overset{?}{ν}$

${}_{6}^{11}C$ $\to$ ${}_{5}^{11}B$ + $e^{+}$ + $ν$

${}_{43}^{97}{T c}$ $\to$ ${}_{42}^{97}{M o}$ + $e^{+}$ + $ν$

${}_{54}^{120}{X e} \to$ ${}_{53}^{120}I$ + $ν$

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