Ncert Solutions Physics Class 12th

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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’ = re

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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 = 3

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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

11 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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

12 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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

Q = -4.0334 Me

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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.

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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) $\times c^{2}$ , where

c = Speed of li

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New answer posted•a year ago

Ncert Solutions Physics Class 12th Nuclei

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 ${}_{54}^{120}{X e}$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ +