Solutions for Exercises of Structure of Atoms and Nuclei

1) Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.

As alpha particle is helium nucleus, it will have 2 protons and 2 neutrons. And we know that

Mass of proton =1.007277amu

Mass of neutron =1.00866amu

Hence, theoretical mass of the alpha particle is,

M_theorotical = (2 × 1.007277)+ (2 × 1.00866)

∴ M_theorotical = 4.031874u

But given actual mass is M = 4.00151 u,

Hence, mass defect is given as,

∆m = 4.031874 − 4.00151

∆m = 0.030364u

We know that 1u mass leads to 931 MeV of energy.

Hence, 0.030364u mass will lead to,

B.E.= 0.030364 × 931 = 28.53MeV

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2) An electron in hydrogen atom stays in its second orbit for 10^-8 s. How many revolutions will it make around the nucleus in that time?

t = 10^-8s for the second orbit in H atom

Number of revolutions = ?

We know that radius of first orbit for hydrogen is

0.053nm i.e. 0.053 × 10^-9m

Hence radius of second orbit is given as,

r² = n²r₁

r² = 4 × 0.053 × 10^-9

r²= 0.212 × 10^-9m

Velocity of electron in any orbit is given as,

$latex v_n\;=\;\frac{nh}{2\pi mr_n}$

$latex v_2\;=\;\frac{2\times6.6\times10^{-34}}{2\pi\times9.1\times10^{-3}\times0.212\times10^{-9}}$

$latex v_2\;=\;1.095\;\times\;10^6m/s$

Hence, time required for one revolution of the second orbit is,

$latex T\;=\;\frac{2\pi r_2}{v_2}$

And hence total number of revolutions made in given time is given as,

$latex =\frac tT=\frac{tv_2}{2\pi r_2}$

$latex =\;\frac{10^{-8}\times1.295\times10^6}{2\times\pi\times0.212\times10^{-9}}$

$latex T\;=\;8.23\;\times\;10^6revs$

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3) Determine the binding energy per nucleon of the americium isotope $latex {}_{95}^{244}Am$, given the mass of $latex {}_{95}^{244}Am$ to be 244.06428 u.

Americium isotope $latex {}_{95}^{244}Am$ will have 95 protons and (244 − 95 = 149) neutrons.

And we know that

Mass of proton =1.007277amu

Mass of neutron =1.00866amu

Hence, theoretical mass of the alpha particle is,

M_theorotical = (95 × 1.007277) + (149 × 1.00866)

∴ M_theorotical = 245.9816u

But given actual mass is

M = 244.06428 u,

Hence, mass defect is given as,

∆m = 245.9816− 244.06428

∆m = 1.9173u

We know that 1u mass leads to 931 MeV of energy.

Hence, 1.9173u mass will lead to,

B.E. = 1.9173 × 931

B.E. = 1785.07MeV

∴ B.E. per Nucleon,

$latex =\;\frac{1785.07}{244}$

$latex =\;7.491\;MeV$

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4) Calculate the energy released in the nuclear reaction $latex {}_3^7Li\;+\;p\;\rightarrow\;2\alpha$ a given mass of Li atom and of helium atom to be 7.016 u and 4.0026 u respectively.

Energy released in the reaction due to the difference in the masses of reactants and products.

Mass of products = 2 × 4.0026 = 8.0052u

Mass of reactants = 7.016 + 1.007277

= 8.023277u

Hence , ∆m = 8.023277− 8.0052

∆m = 0.018077u

We know that 1u mass leads to 931 MeV of energy.

Hence, 0.018077u of mass will lead to,

Energy = 0.018077 × 931

Energy = 16.9789MeV

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5) Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 X 10^-12 per second.

N = 12.3 decay/gram/minute,

N₀ = 15.3 decay/gram/minute

λ = 3.839 × 10^-1 s^-1, t = ?

We know that,

N = N₀e^-λt

$latex \therefore\;12.3\;=\;15.3e^{-(3.839\times10^{-1})t}$

$latex \therefore\;e^{-(3.839\times10^{-1})t}\;=\;\frac{12.3}{15.3}$

$latex \therefore\;e^{(3.839\times10^{-1})t}\;=\;\frac{15.3}{12.3}$

$latex \therefore\;(3.839\times10^{-12})t\;=\;\ln\left(\frac{15.3}{12.3}\right)$

$latex \therefore\;(3.839\times10^{-12})t\;==\;\ln\;(1.2439)$

$latex \therefore\;(3.839\times10^{-12})t=\frac{\log\left(1.2439\right)}{\log\left(2.718\right)}$

$latex \therefore\;(3.839\times10^{-12})t=0.21825$

$latex \therefore\;t\;=\;\frac{0.21825}{3.839\times10^{-12}}$

$latex \therefore\;t\;=\;0.05685\;\times\;10^{12}s$

$latex \therefore\;t\;=\;\frac{0.05685\times10^{12}}{365\times24\times60\times60}\;years$

$latex \therefore\;t\;=\;1803yrs\;\;(Use\;Log)$

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6) The half-life of $latex {}_{38}^{90}Sr$ is 28 years. Determine the disintegration rate of its 5 mg sample.

T_1/2 = 28 years

= 28 × 365 × 24 × 60 × 60 s

dN/dt =?

We know that,

$latex \lambda\;=\;\frac{0.693}{T_{1/2}}$

$latex \lambda\;=\;\frac{0.693}{28\times365\times24\times60\times60}$

$latex \lambda\;=\;7.84\times10^{-10}s^{-1}$

Now 1 mole of Sr i.e. 90 g will contain 6.022×10²³ atoms. Hence 5 mg i.e. 5×10^-3g will contain following number of atoms.

$latex N\;=\;\frac{5\times10^{-3}\times6.022\times10^{23}}{90}$

$latex N\;=\;0.3343\;\times\;10^{20}atoms\;$

Hence rate of disintegration is given as,

$latex \frac{dN}{dt}\;=\;\lambda N$

$latex \therefore\;\frac{dN}{dt}\;=\;7.84\;\times\;10^{-1}\;\times\;0.3343\;\times\;10^{20}$

$latex \therefore\;\frac{dN}{dt}\;=\;2.626\;\times\;10^{10}s^{-1}$

(can use log for calculations)

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7) What is the amount of $latex {}_{27}^{60}Co$ necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?

$latex \frac{dN}{dt}\;=\;10mCi\;$

$latex \frac{dN}{dt}\;=\;10\;\times\;10^{-3}\;\times\;3.7\times10^{10}decay/s,\;$

$latex i.e.\;\frac{dN}{dt}=3.7\times10^8decay/s\;N=?$

$latex T_\frac12=5.3\;yrs$

$latex T_\frac12=\;5.3\;\times\;365\;\times\;24\;\times\;60\;\times\;60\;s$

$latex T_\frac12=\;\;1.67\times\;10^8s\;$

Decay constant is given as,

$latex \lambda\;=\;\frac{0.693}{T_{\displaystyle\frac12}}$

$latex \therefore\;\lambda\;=\;\frac{0.693}{1.67\times10^8}s^{-1}$

Now we know that,

$latex \frac{dN}{dt}\;=\;\lambda N$

$latex \therefore\;N=\frac{\left({\displaystyle\frac{dN}{dt}}\right)}\lambda$

$latex \therefore\;N=\frac{3.7\times10^8\times1.67\times10^8}{0.693}$

$latex \therefore\;N\;=\;8.916\;\times\;10^{16}atoms\;$

Now 1 mole of Co i.e. 60g of Co consists of 6.022 × 10²³ number of atoms. Hence, N atoms corresponds to following mass.

$latex \therefore\;mass\;=\;\frac{60\times8.916\times10^{16}}{6.022\times10^{23}}$

$latex \therefore\;mass\;=\;8.88\;\times\;10^{-6}g\;$

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8) Disintegration rate of a sample is 10¹⁰ per hour at 20 hrs from the start. It reduces to 6.3 x 10⁹ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.

$latex {\left(\frac{dN}{dt}\right)}_{t=20h}\;=\;10^{10}\;per\;hour,$

$latex {\left(\frac{dN}{dt}\right)}_{t=30h}\;=\;6.3\;\times10^9\;per\;hour,$

$latex T_\frac12=?,\;N_0=?$

We know that,

$latex \frac{dN}{dt}\;=\;\lambda N.

$latex But,\;N\;=\;N_0e^{-\lambda t}$

$latex \therefore\;\frac{dN}{dt}\;=\;N_0\lambda e^{-\lambda t}$

$latex \therefore\;10^{10}\;=\;N_0\lambda e^{-\lambda20}\;\;\;\;after\;20\;hrs…(1)$

$latex \&\;6.3\;\times\;10^9\;=\;N_0\lambda e^{-\lambda30}\;\;\;\;after\;30\;hrs…(2)$

$latex \therefore\;\frac{10^{10}}{6.3\times10^9}=\frac{e^{-\lambda20}}{e^{-\lambda30}}=e^{20}$

$latex \therefore\;e^{10\lambda}\;=\;1.5873\;$

$latex \therefore\;10\lambda=\frac{\log\left(1.5873\right)}{\log(2.718)}=0.4620$

∴ λ = 0.4620

$latex \therefore\;T_\frac12=\frac{0.693}\lambda$

$latex =\frac{0.693}{0.04692}=15\;hrs$
( may use log for calculation )

Put λ = 0.0462 in equation (1) we have,

10¹⁰ = N₀(0.0462)e^-(0.0462)20

∴ 10¹⁰ = 0.0462N₀e^-0.924

$latex \therefore\;N_0\;=\;\frac{10^{10}}{0.0462\times e^{-0.924}}$

$latex \therefore\;N_0\;=\;\frac{10^{10}\times e^{0.924}}{0.0462}$

$latex \therefore\;N_0\;=\;\frac{10^{10}\times2.5188}{0.0462}$

$latex \therefore\;N_0\;=\;5.45\;\times\;10^{11}atoms$

[e^0.924 ⇒ Let e^0.924 = x, taking log both sides,

0.924log_e(e) = log_ex, converting to base 10,

$latex 0.924\;\times\;1\;=\frac{\log(x)}{\log(2.718)}\\$

$latex \therefore\;\log(x)\;=\;0.924\;\times\;0.4342\\$

$latex \therefore\;x\;=\;anti\log(0.4012)\;=\;2.5188\\$

$latex \therefore\;e^{0.924}\;=\;2.5188\;\\$

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9) The isotope ⁵⁷Co decays by electron capture to ⁵⁷Fe with a half-life of 272 d. The ⁵⁷Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays. (a) Find the mean lifetime and decay constant for ⁵⁷Co. (b) If the activity of a radiation source ⁵⁷Co is 2.0 µCi now, how many ⁵⁷Co nuclei does the source contain? (c) What will be the activity after one year?

⁵⁷Co → ⁵⁷Fe + e⁺

T_1/2 = 272 d

= 272 × 24 × 60 × 60 s

= 2.35 × 10⁷s

a) Decay constant is given as,

$latex \lambda\;=\;\frac{0.693}{T_{1/2}}\\$

$latex \lambda\;=\;\frac{0.693}{2.35\times10^7}\\$

$latex \lambda\;=\;2.95\;\times10^{-8}s^{-1}\\$

Mean lifetime is given as,

$latex \tau\;=\;\frac1\lambda\\$

$latex \therefore\;\tau\;=\;\frac1{2.95\times10^{-8}}\\$

$latex \therefore\;\tau\;=\;3.39\;\times\;10^7s\;\\$

$latex b)\;\frac{dN}{dt}\;=\;2\mu Ci\\$

$latex =\;2\;\times\;10^{-6}\;\times\;3.7\;\times\;10^{10}decay/s\\$

$latex \therefore\;\frac{dN}{dt}\;=\;7.40\;\times\;10^4decay/s\;,N=?\\$

$latex also,\;\frac{dN}{dt}\;=\;\lambda N\\$

$latex \therefore\;N\;=\;\;\frac{dN/dt}\lambda\;\\$

$latex \therefore\;N\;=\;\;\frac{7.40\times10^4}{2.95\times10^{-8}}\\$

$latex \therefore\;N\;=\;\;2.51\;\times\;10^{12}\;nuclei\\$

$latex c)\;\frac{dN}{dt}\;=\;?\;after\;1\;yr,\\$

Original activity = 2μCi

= 3.7 × 10¹⁰decay/s

$latex {\left(\frac{dN}{dt}\right)}_{original}\;=\;2\mu Ci\\$

= 3.7 × 10¹⁰decay/s

We know that,

N = N₀e^-λt

Hence after 1 year, we have.

N = N₀e^{-(2.95×10-8×365×24×60×60)}

N = N₀e^-(0.9303)

∴ N = 0.394N₀

(Find e^-(0.9303) using method discussed above)

Hence activity is given as,

$latex \left(\frac{dN}{dt}\right)\;=\;\lambda N\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{original}=\;\lambda N_0\;\&\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;yr}=\;\lambda N\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;year}=\;\lambda\left(0.394N_0\right)\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;year}=\;\lambda N_00.394\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;year}=\;{\left(\frac{dN}{dt}\right)}_{original}\times0.394\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;year}=\;2\mu Ci\;\times\;0.394\\$

$latex \therefore\;{\left(\frac{dN}{dt}\right)}_{1\;year}=\;0.789\;\mu Ci\\$

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10) A source contains two species of phosphorous nuclei,$latex {}_{15}^{32}P\left(T_{1/2}=14.3\;d\right)\\$ and $latex {}_{15}^{33}P\left(T_{1/2}=25.3\;d\right)\\$. At time t = 0, 90% of the decays are from $latex {}_{15}^{32}P\\$. How much time has to elapse for only 15% of the decays to be from $latex {}_{15}^{32}P\\$ ?

We know that,

$latex \frac{dN}{dt}\;\propto\;N$

Hence, initially as 90% of the decays are from $latex {}_{15}^{32}P\\$. we have, at t = 0, ratio of $latex {}_{15}^{32}P\\$ and $latex {}_{15}^{33}P\\$ nuclei is 9:1. That is $latex {}_{15}^{32}P\\$ nuclei are 9x while $latex {}_{15}^{33}P\\$ nuclei are x.

Finally we want time t such that 15% of the decays to be from $latex {}_{15}^{32}P\\$. That is ratio of $latex {}_{15}^{32}P\\$ to $latex {}_{15}^{33}P\\$ nuclei should become 15:85 i.e. 3:17. That is $latex {}_{15}^{32}P\\$ nuclei are 3y while $latex {}_{15}^{33}P\\$ nuclei are 17y.

We know that,

$latex N\;=\;\frac{N_0}{2^n}$ here n is the number of half life period in given time.

$latex \;i.e.\;n\;=\frac t{T_{1/2}}$

$latex \therefore\;N\;=\frac{N_0}{2^n}=\frac{N_0}{2^{\displaystyle\frac t{T_{1/2}}}}$

Hence for $latex {}_{15}^{32}P\\$ we have,

$latex 3y\;=\;\frac{9x}{2^{t/14.3}}$

And for $latex {}_{15}^{33}P\\$ we have,

$latex 17y\;=\;\frac x{2^{t/25.3}}$

$latex Also\;\frac{3y}{17y}$

$latex =\;\frac{\displaystyle\frac{9x}{2^{t/14.3}}}{\displaystyle\frac x{2^{t/25.3}}}$

$latex =\;\frac{9x}{2^{t/14.3}}\;\times\;\frac{2^{t/25.3}}x$

$latex \therefore\;\frac3{17\times9}=\frac{2^{t/25.3}}{2^{t/14.3}}$

$latex =\;2^{\frac t{25.3}-\frac t{14.3}}$

$latex =\;2^{-t\left(\frac1{14.3}-\frac1{25.3}\right)}$

$latex =\;2^{-0.0304t}$

$latex \therefore\;2^{0.0304}=\frac{17\times9}3\;=\;51$

$latex \therefore\;\log\left(2^{0.0304}\right)=\log51$

$latex \therefore\;0.0304t\;\times\;\log2\;=\;1.7075$

$latex \therefore\;0.0304t\;=\;\frac{1.7075}{0.3010}$

$latex \therefore t\;=\;\frac{5.672}{0.0304}$

$latex \therefore t\;=\;186.6\;d$

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11) How much mass of ²³⁵U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV

Energy per second = 3000 × 10⁶ = 3 × 10⁹J

Average Energy per fission = 202.79MeV

= 202.79 × 10⁶ × 1.6 × 10^-19

= 3.244× 10^-1 J

Hence number of fission or number of atoms required per second,

$latex =\;\frac{3\;\times\;10^9}{3.244\;\times\;10^{-11}}$

$latex =\;9.2478\;\times\;10^{19}$

Hence number of fission or number of atoms each day,

= 9.2478 × 10¹⁹ × 60 × 60 × 24 = 8 × 10²⁴

But 1 mole of substance (235g) will contain 6.022 × 10²³ atoms. Hence mass of (8 × 10²⁴) atoms is given as,

$latex m\;=\;\frac{\left(8\;\times\;10^{24}\right)}{6.022\;\times\;10^{23}}\;\times\;235$

$latex \therefore\;m\;=\;3121.89g\;=\;3.1kg\;$

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12) In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are $latex {}_{12}^{24}Mg\;(23.98504\;u)$, $latex {}_{12}^{25}Mg\;(24.98584\;u)$ and $latex {}_{12}^{26}Mg\;(25.98259\;u)$. The natural abundance of $latex {}_{12}^{24}Mg$ is 78.99% by mass. Calculate the abundances of other two isotopes.

Let the abundance of $latex {}_{12}^{25}Mg$ by mass is x%.

Therefore, the abundance of $latex {}_{12}^{26}Mg$ by mass will be equal to (100 − 78.99 − x) = (21.01 − x)%

Average atomic mass of the magnesium is given as weighted average of its isotopes.

$latex \therefore\;24.312=23.98504\times\frac{78.99}{100}+24.98584\times\frac x{100}$

$latex \therefore\;24.312=23.98504\times\frac{78.99}{100}+24.98584\times\frac x{100}+25.98259\times\frac{\left(21.01-x\right)}{100}$

Solving we get, x = 9.303.

Hence abundance of $latex {}_{12}^{25}Mg$ by mass is 9.303%.

And the abundance of $latex {}_{12}^{26}Mg$ by mass will be equal to (21.01 − 9.303) = 11.71%

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