Nuclei

13.17 The average energy released per fission of ${}_{94}{}^{239}Pu$  $E_{avg}$= 180 MeV

Amount of pure${}_{94}{}^{239}u$ , m = 1 kg = 1000 g

Avogadro's number,
$N_A$ = 6.023 $*{10}^{23}$

Mass number of ${}_{94}{}^{239}\mathit{P}\mathit{u}$  = 239 gm

Hence, number of atoms in 1000 g ${}_{94}{}^{239}Pu$, N =$\frac{6.023*{10}^{23}}{239}$ $*1000$= 2.52 $*{10}^{24}$

Total energy released during the fission of 1 kg of  ${}_{94}{}^{239}Pu$, E =$E_{avg}$ $*$N

= 180 $*$ 

13.16 The fission can be shown as:

${}_{26}{}^{56}Fe\to 2{}_{13}{}^{28}Al$ 

It is given that atomic mass

m ( ${}_{26}{}^{56}Fe)$= 55.93494 u

m ( ${}_{13}{}^{28}Al)$= 27.98191 u

The Q-value of this reaction is given as:

Q = $\left[m\left({}_{26}{}^{56}Fe\right)-2m\left({}_{13}{}^{28}Al\right)\right]c^2$

= $\left[55.93494\mathrm{}-2\times 27.98191\right]c^2$

= -0.02888 $c^2$

= -0.02888$\times 931.5$  MeV

= - 26.902 MeV

The Q value of the fission is negative, therefore the fission is not possible energetically. Q value needs t

13.14 In ${\beta }^{-}$ emission, the number of protons increase by 1 and one electron and an antineutrino are emitted from the parent nucleus.

${\beta }^{-}$ emission of the nucleus ${}_{10}{}^{23}Ne$ :

${}_{10}{}^{23}Ne$ $\to {}_{11}{}^{23}Na$ + $e^{-}$ + $\stackrel{?}{\nu }$ + Q

It is given that:

Atomic mass m ( ${}_{10}{}^{23}Ne)$ = 22.994466 u

Atomic mass m (&

13.11 Nuclear radius of the gold isotope,  ${}_{47}{}^{197}Au$ = $R_{Au}$

Nuclear radius of silver isotope,  ${}_{47}{}^{107}Ag$ = $R_{Ag}$

Mass number of gold,  $A_{Au}$ = 197

Mass number of silver,  $A_{Ag}$ = 107

The ratio of the radii of the two nuclei is related with their mass numbers as :

$\frac{R_{Au}}{R_{Ag}}$ = ${\left. [\frac{A_{Au}}{A_{Ag}}\right]}^{1/3}$ = ${\left. [\frac{197}{107}\right]}^{1/3}$&n

13.10 Half life of ${}_{38}{}^{90}Sr$ , $T_{1/2}$ = 28 years = 28 $\times 365\times 24\times 60\times 60$ secs = 0.883 $\times {10}^9$ s

Mass of the isotope, m = 15 mg = 15 $\times {10}^{-3}$ gms

90 g of ${}_{38}{}^{90}Sr$ contains 6.023 $\times {10}^{23}$ atoms

No. of atoms in 15 mg of ${}_{38}{}^{90}Sr$ contains = $\frac{6.023\times {10}^{23}}{90}\times$ 15 $\times {10}^{-3}$ = 1.0038 $\times {10}^{20}$

Rate of disintegrat

13.9 The strength of the radioactive source is given as:

$\frac{dN}{dt}$ = 8.0 mCi = 8 $\times {10}^{-3}$ $\times 3.7\times {10}^{10}$ decay/s = 296 $\times {10}^6$ decay/s, where

N = Required number of atoms

Given, half life of ${}_{27}{}^{60}Co$ , $T_{1/2}$ = 5.3 years = 5.3 $\times 365\times 24\times 60\times 60$ secs = 167 $\times {10}^6$ s

For decay constant $\lambda$ , we have rate

13.8 Decay rate of living carbon-containing matter, R = 15 decay / min

Half life of ${}_6{}^{14}C$ , $T_{1/2}$ = 5730 years

Decay rate of the specimen obtained from the Mohenjo-Daro site, R’ = 9 decays/min

Let N be the number of radioactive atoms present in a normal carbon-containing matter.

Let N’ be the

13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope = $N_o$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of $N_o$ remains after decay. Hence, we can write,

$\frac{N}{N_o}$ = 3.125% = $\frac{3.125}{100}$ = $\frac{1}{32}$

But $\frac{N}{N_o}$ = $e^{-\lambda t}$ ,

13.5 Mass of the copper coin, m’ = 3.0 g

Atomic mass of ${}_{29}{}^{63}Cu$ , m = 62.92960 u

The total number of ${}_{29}{}^{63}Cu$ atoms in the coin, N = $\frac{N_A\times m\text{'}}{Massnumber}$ , where

$N_A$ = Avogadro’s number = 6.023 $\times {10}^{23}$ atoms / g

Mass number = 63 g

Therefore, N = $\frac{6.023\mathrm{}\times {10}^{23}\times 3}{63}$ = 2.868 $\times {10}^{22}$ atoms

${}_{29}{}^{63}Cu$ has 29 protons and

13.3 Atomic mass of ${}_7{}^{14}N$ nitrogen , m = 14.00307 u

A nucleus of ${}_7{}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7 $m_p$ + 7 $m_n$ - m, where

Mass of a proton, $m_p$ = 1.007825 u

Mass of a neutron, $m_n$ = 1.008665 u

Therefore,