5 Marks Questions - Physics STD 11 Science Questions - Vidyadip

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

If the nucleons of a nucleus are separated from each other, the total mass is increased. Where does this mass come from?

Answer

When the nucleons of a nucleus are separated, a certain amount of energy is to be given to the nucleus, which is known as the binding energy. Binding energy = [(Number of nucleons) × (Mass of a nucleon) - (Mass of the nucleus)] When the nucleons of a nucleus are separated, the increase in the total mass comes from the binding energy, which is given to the nucleus to break-off its constituent nucleons as energy is related to mass by the relation given below.$\text{E}=\Delta\text{mc}^2$

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Question 25 Marks

$4\times10^{23}$ tritium atoms are contained in a vessel. The half-life of decay tritium nuclei is 12.3y. Find:

The activity of the sample.
The number of decay in the next 10 hours.
The number of decays in the next 6.15y.

Answer

$\text{N}=4\times10^{23};\text{t}_{\frac{1}{2}}=12.3\text{ years}.$

Activity $=\frac{\text{dN}}{\text{dt}}=\lambda\text{n}=\frac{0.693}{\text{t}_{\frac{1}{2}}}\text{N}=\frac{0.693}{12.3}\times4\times10^{23}\text{dis/year}.$

$=7.146\times10^{14}\text{dis/sec}.$

$\frac{\text{dN}}{\text{dt}}=7.146\times10^{14}$

No. of decays in next 10 hours $=7.146\times10^{14}\times10\times36$

$=257.256\times10^{17}=2.57\times10^{19}$

$\text{N = N}_0\text{e}^{-\lambda\text{t}}=4\times10^{23}\times\text{e}^{\frac{-0.693}{20.3}\times6.16}\\=2.82\times10^{23}$ = No. of atoms remained

No. of atoms disintegrated $=(4-2.82)\times10^{23}=1.18\times10^{23}$

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Question 35 Marks

${ }^{238} \mathrm{U}$ decays to ${ }^{206} \mathrm{~Pb}$ with a half-life of $4.47 \times 10^9 \mathrm{y}$. This happens in a number of steps. Can you justify a single half for this chain of processes? A sample of rock is found to contain 2.00 mg of ${ }^{238} \mathrm{U}$ and 0.600 mg of ${ }^{206} \mathrm{~Pb}$. Assuming that all the lead has come from uranium, find the life of the rock.

Answer

Half life period can be a single for all the process. It is the time taken for $\frac{1}{2}$ of the uranium to convert to lead. No. of atoms of $U^{238}=\frac{6\times10^{23}\times2\times10^{-3}}{238}=\frac{2}{238}\times10^{20}=0.05042\times10^{20}$ No. of atoms in Pb $=\frac{6\times10^{23}\times0.6\times10^{-3}}{206}=\frac{3.6}{206}\times10^{20}$ Initially total no. of uranium atoms $=\Big(\frac{12}{235}+\frac{3.6}{206}\Big)\times10^{20}=0.06789$$\text{N = N}_0\text{e}^{-\lambda\text{t}}\Rightarrow\text{N = N}_0\text{e}^{\frac{-0.693}{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}$
$\Rightarrow0.05042=0.06789\text{e}^{\frac{-0.693}{4.47\times10^9}}$
$\Rightarrow\log\Big(\frac{0.05042}{0.06789}\Big)=\frac{-0.693\text{t}}{4.47\times10^9}$
$\Rightarrow\text{t}=1.92\times10^9\text{ years}.$

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Question 45 Marks

Calculate the Q-values of the following fusion reactions:

$\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_1\text{H}+\text{ }^1_1\text{H}$
$\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_2\text{H}+\text{n}$
$\text{ }^2_1\text{H}+\text{ }^3_1\text{H}\rightarrow\text{ }^4_1\text{H}+\text{n}$

Atomic masses are$\text{m}\big(\text{ }^2_1\text{H}\big)=2.014102\text{u}$
$\text{m}\big(\text{ }^3_1\text{H}\big)=3·016049\text{u}$
$\text{m}\big(\text{ }^3_2\text{H}\big)=3.016029\text{u}$
$\text{m}\big(\text{ }^4_2\text{He}\big)=4·002603\text{u}$

Answer

$\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_1\text{H}+\text{ }^1_1\text{H}$

Q value $=2\text{M}\big(\text{ }^2_1\text{H}\big)=\big[\text{M}\big(\text{ }^3_1\text{H}\big)+\text{M}\big(\text{ }^{3}_1\text{H}\big)\big]$

$=[2\times2.014102-(3.016049 + 1.007825)]\text{u}$

$= 4.0275\text{Mev} = 4.05\text{Mev.}$

$\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_2\text{H}+\text{n}$

Q value $=2\big[\text{M}\big(\text{ }^2_1\text{H}\big)-\text{M}\big(\text{ }^3_2\text{He}\big)+\text{M}_{\text{n}}\big]$

$=[2\times2.014102 -(3.016049 + 1.008665)]\text{u}$

$=3.26\text{Mev}= 3.25\text{Mev.}$

$\text{ }^2_1\text{H}+\text{ }^3_1\text{H}\rightarrow\text{ }^4_1\text{H}+\text{n}$

Q value $=\big[\text{M}\big(\text{ }^2_1\text{H}\big)+\text{M}\big(\text{ }^3_1\text{He}\big)-\text{M}\big(\text{ }^4_2\text{He}\big)+\text{M}_{\text{n}}\big]$

$=(2.014102 + 3.016049)-(4.002603 + 1.008665)]\text{u}$

$=17.58\text{Mev} = 17.57\text{Mev.}$

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Question 55 Marks

A point source emitting alpha particles is placed at a distance of $1$ m from a counter which records any alpha particle falling on its $1 cm^2$ window. If the source contains $6.0 \times 10^{16}$ active nuclei and the counter records a rate of $50000$ counts/ second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.

Answer

Counts received per $cm^2 = 50000$ Counts/sec. $N = N_3o$ of active nucleic $= 6 \times 10^{16}$ Total counts radiated from the source = Total surface area $\times 50000 \ counts/cm^2 = 4 \times 3.14 \times 1 \times 10^4 \times 5 \times 10^4 = 6.28 \times 10^9$ $\text{Counts}=\frac{\text{dN}}{\text{dt}}$ We know, $\frac{\text{dN}}{\text{dt}}=\lambda\text{N}$$\lambda=\frac{6.28\times10^9}{6\times10^{16}}=1.0467\times10^{-7}=1.05\times10^{-7}\text{s}^{-1}$

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Question 65 Marks

Natural water contains a small amount of tritium $\big(\text{ }^3_1\text{H}\big).$ This isotope beta-decays with a half-life of $12.5$ years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottled of whisky. On returning, he analyses the whisky and finds that it contains only $1.5$ per cent of the $\text{}^3_1\text{H}$ radioactivity as compared to a recently purchased bottle marked '$8$ years old'. Estimate the time of that unsuccessful attempt.

Answer

The activity when the bottle was manufactured $= A_0$ Activity after 8 years $=\text{A}_0\text{e}^{\frac{-0.693}{12.5}}\times8$ Let the time of the mountaineering = t years from the present$\text{A = A}_0\text{e}^{\frac{-0.693}{12.5}\times\text{t}};$ A = Activity of the bottle found on the mountain.
A = (Activity of the bottle manufactured 8 years before) × 1.5%$\Rightarrow\text{A}_0\text{e}^{\frac{-0.693}{12.5}}=\text{A}_0\text{e}^{\frac{-0.693}{12.5}\times8}\times0.015$
$\Rightarrow\frac{-0.693}{12.5}\text{t}=\frac{-0.693\times8}{12.5}+\text{In}[0.015]$
$\Rightarrow0.05544\text{t = 0.44352 + 4.1997}\Rightarrow\text{t}=83.75\text{ years}.$

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Question 75 Marks

$\text{ }^{212}_{33}\text{Bi}$ can disintegrate either by emitting an $\alpha$-particle of by emitting a $\beta^-$-particle.

Write the two equations showing the products of the decays.
The probabilities of disintegration $\alpha$ and $\beta$-decays are in the ratio $\frac{7}{13}$. The overall half-life of ${ }^{212} Bi$ is one hour. If 1 g of pure ${ }^{212} Bi$ is taken at 12.00 noon, what will be the composition of this sample at $1 P . m$. the same day?

Answer

$\text{ }^{212}_{83}\text{Bi}\rightarrow\text{ }^{208}_{81}\text{Ti}+\text{ }^{4}_2\text{He}(\alpha)$$\text{ }^{212}_{83}\text{Bi}\rightarrow\text{ }^{212}_{84}\text{Bi}\rightarrow\text{ }^{212}_{84}\text{P}_0+\text{e}^-$
$\text{t}_{\frac{1}{2}}=1\text{h}.$ Time elapsed = 1 hour
at $t =0 Bi ^{212}$ Present $=1 g$
$\therefore$ at $t =1 Bi ^{212}$ Present $=0.5 g$
Probability $\alpha$-decay and $\beta$-decay are in ratio $\frac{7}{13}$.
$\therefore$ TI remained $=0.175 g$
$\therefore P_0$ remained $=0.325 g$

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Question 85 Marks

A town has a population of 1 million. The average electric power needed per person is 300 W . A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at $25 \%$.
a. Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day.
b. Assuming the fission to take place largely through ${ }^{235} U$, at what rate will the amount of ${ }^{235} U$ decrease? Express your answer in kg per day.
c. Assuming that uranium enriched to $3 \%$ in ${ }^{235} U$ will be used, how much uranium is needed per month ( 30 days)?

Answer

Energy radiated per fission $=2\times10^8\text{ev}$

Usable energy $=2\times10^8\times\frac{25}{100}=5\times10^7\text{ev}=5\times1.6\times10^{-12}$
Total energy needed $300\times10^8 =3\times10^8 \text{J/s}$
No. of fission per second $=\frac{3\times10^8}{5\times1.6\times10^{-12}}=0.375\times10^{20}$
No. of fission per day $=0.375\times10^{20}\times3600\times24 = 3.24\times10^{24}\text{ fissions}$

From ‘a’ No. of atoms disintegrated per day $= 3.24 \times 10^{24}$

We have, $6.023 \times 10^{23}$ atoms for 235g
for $3.24 × 10^{24} \text{ atom}=\frac{235}{6.023\times10^{23}}\times3.24\times10^{24}\text{g}$
$=1264\text{g/day}=1.264\text{kg/day}$

Total uranium needed per month = 1.264 × 30kg

Let x kg of uranium enriched to 3% in $^{235}U$ be used.
$\Rightarrow\text{x}\times\frac{3}{100}=1.264\times30$
$\Rightarrow\text{x}=1264\text{kg}$

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Question 95 Marks

The half-life of ${ }^{40} \mathrm{~K}$ is $1.30 \times 10^9 \mathrm{y}$. A sample of 1.00 g of pure KCl gives 160 counts/s. Calculate the relative abundance of ${ }^{40} \mathrm{~K}$ (fraction of ${ }^{40} \mathrm{~K}$ present) in natural potassium.

Answer

Given: Half life period $\text{t}_{\frac{1}{2}}=1.30\times10^9\text{year}$$\text{A}=160\text{ count/s}=1.30\times10^9\times365\times86400$
$\therefore\text{A}=\lambda\text{N}\Rightarrow160=\frac{0.693}{\text{t}_{\frac{1}{2}}}\text{N}$
$\Rightarrow\text{N}=\frac{160\times1.30\times365\times86400\times10^9}{0.693}=9.5\times10^{18}$
$\therefore6.023\times10^{23}$ No. of present in 40 grams.
$6.023\times10^{23}=40\text{g}\Rightarrow1=\frac{40}{6.023\times10^{23}}$
$\therefore9.5\times10^{18}$ present in $=\frac{40\times9.5\times10^{18}}{6.023\times10^{23}}=6.309\times10^{-4}=0.00063$
$\therefore$ The relative abundance at 40k in natural potassium $=(2 \times 0.00063 \times 100)\%=0.12\%$

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Question 105 Marks

In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged?

Answer

In beta decay, a neutron from the nucleus is converted to a proton releasing an electron and an antineutrino or a proton is converted to a neutron releasing a positron and a neutrino.$\beta^-\text{decay:}\text{ n}\rightarrow\text{p + e + }\vec{\text{v}}$
$\beta^+\text{decay:}\text{ p}\rightarrow\text{n + e}^+ + \text{v}$
Since the number of valence electrons present in the parent atom do not change, the remaining atom does not get oppositely charged. Instead, due to a change in the atomic number, there's a formation of a new element.

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Question 115 Marks

A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is $4.0 \times 10^{30} \mathrm{~kg}$ (twice the mass of the sun).

Answer

$\text{f}=\frac{\text{M}}{\text{v}}\Rightarrow\text{V}=\frac{\text{M}}{\text{f}}=\frac{4\times10^{30}}{2.4\times10^{17}}$$=\frac{1}{0.6}\times10^{13}=\frac{1}{6}\times10^{14}$
$\text{V}=\frac{4}{3}\pi\text{R}^3.$
$\Rightarrow\frac{1}{6}\times10^{14}=\frac{4}{3}\pi\times\text{R}^3$
$\Rightarrow\text{R}^3=\frac{1}{6}\times\frac{3}{4}\times\frac{1}{\pi}\times10^{14}$
$\Rightarrow\text{R}^3=\frac{1}{8}\times\frac{100}{\pi}\times10^{12}$
$\therefore\text{R}=\frac{1}2{}\times10^4\times3.17=1.585\times10^4\text{m}=15\text{km}.$

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Question 125 Marks

What is the difference between cathode rays and beta rays? When the two are travelling in space, can you make out which is the cathode ray and which is the beta ray?

Answer

Cathode rays consist of electrons that are accelerated using electrodes. They do not carry high energy and do not harm human body. On the other hand, beta rays consist of highly energetic electrons that can even penetrate and damage human cells. Beta rays are produced by the decay of radioactive nuclei. If the two are travelling in space, they can be distinguished by the phenomenon named production of Bremsstrahlung radiation, which is produced by the deceleration of a high energy particle when deflected by another charged particle, leading to the emission of blue light. Only beta rays are capable of producing it.

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Question 135 Marks

In an agricultural experiment, a solution containing 1 mole of a radioactive material $\Big(\text{t}_{\frac{1}{2}}=14.3\text{ days}\Big)$ was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruit. If the activity measured was $1\mu\text{Ci},$ what per cent of activity is transmitted from the root to the fruit in steady state?

Answer

$\text{n}=1\text{ mole}=6\times10^{23}\text{ atoms},\text{ t}_{\frac{1}{2}}=14.3\text{ days}$$\text{t}=70\text{ hours},\frac{\text{dN}}{\text{dt}}$ in root after time $\text{t}=\lambda\text{N}$
$\text{N = No e}^{-\lambda\text{t}}=6\times10^{23}\times\text{e}^{\frac{-0.693\times70}{14.3\times24}}$
$=6\times10^{23}\times0.868=5.209\times10^{23}$
$5.209\times10^{23}\times\frac{-0.693}{14.3\times24}=\frac{0.0105\times10^{23}}{3600}\text{dis/hour.}$
Fraction of activity transmitted $=\Big(\frac{1\mu\text{ci}}{2.9\times10^{17}}\Big)\times100\%$
$\Rightarrow\Big(\frac{1\times3.7\times10^8}{2.9\times10^{11}}\times100\Big)\%=1.275\times10^{-11}\%$

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Question 145 Marks

A vessel of volume $125cm^3$ contains tritium $\big(\text{ }^3\text{H,t}_{\frac{1}{2}}=12.3\text{y}\big)$ at 500 kPa and $300K$. Calculate the activity of the gas.

Answer

$\text{V}=125\text{cm}^3=0.125\text{L, P}=500\text{kPa}=5\text{atm}.$$\text{T}=300\text{K},\text{t}_{\frac{1}{2}}=12.3\text{ years}$
$=3.82\times10^8\text{sec. Activity}=\lambda\times\text{N}$
$\text{N = n}\times6.023\times10^{23}=\frac{5\times0.125}{8.2\times10^{-2}\times3\times10^2}\\\times6.023\times10^{23}=1.5\times10^{22}\text{ atoms}.$
$\lambda=\frac{0.693}{3.82\times10^8}=0.1814\times10^{-8}=1.81\times10^{-9}\text{s}^{-1}$
Activit y $=\lambda\text{N}=1.81\times10^{-9}\times1.5\times10^{22}=2.7\times10^3$ disintegration/sec
$=\frac{2.7\times10^{13}}{3.7\times10^{10}}\text{Ci}=729\text{Ci}$

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Question 155 Marks

If neutrons exert only attractive force, why don't we have a nucleus containing neutrons alone?

Answer

Nuclear forces are short range strong attractive forces that act between two proton-proton, neutron-proton and neutron-neutron pairs. Two protons have strong nuclear forces between them and also exert electrostatic repulsion on each other. However, electrostatic forces are long ranged and have very less effect as compared to the strong nuclear forces.
So, in a nucleus (that is very small in dimension), there's no such significance of repulsive force as compared to the strong attractive nuclear force. On the other hand, an atom contains electrons revolving around its nucleus. These electrons are kept in their orbit by the strong electrostatic force that is exerted on them by the protons present inside the nucleus. Hence, a nucleus contains protons as well as neutrons.

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Question 165 Marks

The half-life of $^{199}Au$ is 2.7 days.

Find the activity of a sample containing $1.00\mu\text{g}$ of $^{198}Au$.
What will be the activity after $7$ days? Take the atomic weight of $^{198}Au$ to be $198g/mol$.

Answer

198 grams of Ag contains → $N_0$ atoms.

$1\mu\text{g of Ag contains}\rightarrow\frac{\text{N}_0}{198}\times1\mu\text{g}=\frac{6\times10^{23}\times1\times10^{-6}}{198}$ atoms

Activity $=\lambda\text{N}=\frac{0.963}{\text{t}_{\frac{1}{2}}}\times\text{N}=\frac{0.693\times6\times10^{17}}{198\times2.7}$ disintegrations/day.

$=\frac{0.693\times6\times10^{17}}{198\times2.7\times3600\times24}$ disintegration/sec

$=\frac{0.693\times6\times10^{17}}{198\times2.7\times36\times24\times3.7\times10^{10}}$ curie = 0.244 Curie.

$\text{A}=\frac{\text{A}_0}{2\text{t}_{\frac{1}{2}}}=\frac{0.244}{2\times\frac{7}{2.7}}=0.0405=0.040$ Curie.

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Question 175 Marks

The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing $^{128}I$ varies with time as follows.

Time t(minute):	$0$	$25$	$50$	$75$	$100$
Ctount rate $R(10^9s^{-1}):$	$30$	$16$	$8.0$	$3.8$	$2.0$

Plot In $\Big(\frac{\text{R}_0}{\text{R}}\Big)$ against t.
From the slope of the best straight line through the points, find the decay constant $\lambda.$
Calculate the half-life $\text{t}_{\frac{1}{2}}.$

Answer

Here we should take $R_0$ at time is $t_0 = 30 \times 10^9s^{-1}$

$\text{ln}\Big(\frac{\text{R}_0}{\text{R}_1}\Big)=\text{ln}\Big(\frac{30\times10^9}{30\times10^9}\Big)=0$
$\text{ln}\Big(\frac{\text{R}_0}{\text{R}_2}\Big)=\text{ln}\Big(\frac{30\times10^9}{16\times10^9}\Big)=0.63$
$\text{ln}\Big(\frac{\text{R}_0}{\text{R}_3}\Big)=\text{ln}\Big(\frac{30\times10^9}{8\times10^9}\Big)=1.35$
$\text{ln}\Big(\frac{\text{R}_0}{\text{R}_4}\Big)=\text{ln}\Big(\frac{30\times10^9}{3.8\times10^9}\Big)=2.06$
$\text{ln}\Big(\frac{\text{R}_0}{\text{R}_5}\Big)=\text{ln}\Big(\frac{30\times10^9}{2\times10^9}\Big)=2.7$

$\therefore$ The decay constant $\lambda=0.028\text{min}^{-1}$
$\therefore$ The half life period $\text{t}_{\frac{1}{2}}$

$\text{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{0.693}{0.028}=25\text{min}.$

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Question 185 Marks

Potassium-$40$ can decay in three modes. It can decay by $\beta^-$-emission, $\beta^+$-emission of electron capture.

Write the equations showing the end products.
Find the Q-values in each of the three cases. Atomic masses of $\text{ }^{40}_{18}\text{Ar},\text{ }^{40}_{19}\text{K}$ and $\text{ }^{40}_{20}\text{Ca}$ are $39.9624\ u$ $39.9640\ u$ and $39.9626u$ respectively.

Answer

Decay of potassium-40 by $\beta^-$ emission is given by

$\text{ }_{19}\text{K}^{40}\rightarrow\text{ }_{20}\text{Ca}^{40}+\beta^-+\bar{\text{v}}$
Decay of potassium-40 by $\beta^+$ emission is given by
$\text{ }_{19}\text{K}^{40}\rightarrow\text{ }_{18}\text{Ar}^{40}+\beta^++\text{v}$
Decay of potassium-40 by electron capture is given by
$\text{ }_{19}\text{K}^{40}+\text{e}^-\rightarrow\text{ }_{18}\text{Ar}^{40}+\text{v}$

$Q _{\text {value }}$ in the $\beta^{-}$decay is given by
$Q_{\text {value }}=\left[m\left(19 K^{40}\right)-m\left(20 Ca^{40}\right)\right] c^2$
$=[39.9640 u-39.9626 u] c^2$
$=0.0014 \times 931 MeV$
$=1.3034 MeV$
$Q _{\text {value }}$ in the $\beta^{+}$decay is given by
$Q_{\text {value }}=\left[m\left({ }_{19} K^{40}\right)-m\left(20 Ar{ }^{40}\right)-2 m_{e}\right] c^2$
$=[39.9640 u-39.9624 u-0.0021944 u] c^2$
$=(39.9640-39.9624) 931 MeV-1022 keV$
$=1489.96 keV-1022 keV$
$=0.4679 MeV$
$Q _{\text {value }}$ in the electron capture is given by
$Q_{\text {value }}=\left[m\left(19 K^{40}\right)-m\left(20 Ar^{40}\right)\right] c^2$
$=(39.9640-39.9624) uc^2$
$=1.4890=1.49 MeV$

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Question 195 Marks

A sample contains a mixture of $^{108}Ag$ and $^{110}Ag$ isotopes each having an activity of $8.0 \times 10^8$ disintegration per second. $^{110}Ag$ is known to have larger half-life than $^{108}Ag$. The activity A is measured as a function of time and the following data are obtained.

Time (s)	Activity (A) ($10^8$ disinte- grations $s^{-1}$)	Time (s)	Activity (A) ($10^8$ disinte-grations $s^{-1}$)
20	11.799	200	3.0828
40	9.1680	300	1.8899
60	7.4492	400	1.1671
80	6.2684	500	0.7212
100	5.4115

Plot ln $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time.
See that for large values of time, the plot is nearly linear. Deduce the half-life of $^{110}Ag$ from this portion of the plot.
Use the half-life of $^{110}Ag$ to calculate the activity corresponding to $^{108}Ag$ in the first 50s.
Plot In $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time for $^{108}Ag$ for the first 50s.
Find the half-life of $^{108}Ag$.

Answer

Activities of sample containing $^{108}Ag$ and $^{110}Ag$ isotopes = $8.0 \times 10^8$ disintegration/sec.

Here we take A = 8 × 108 dis./sec

$\text{ln}\Big(\frac{\text{A}_1}{\text{A}_{0_{1}}}\Big)=\text{ln}\Big(\frac{11.79}{8}\Big)=0.389$
$\text{ln}\Big(\frac{\text{A}_2}{\text{A}_{0_{2}}}\Big)=\text{ln}\Big(\frac{9.1680}{8}\Big)=0.1362$
$\text{ln}\Big(\frac{\text{A}_3}{\text{A}_{0_{3}}}\Big)=\text{ln}\Big(\frac{7.4492}{8}\Big)=-0.072$
$\text{ln}\Big(\frac{\text{A}_4}{\text{A}_{0_{4}}}\Big)=\text{ln}\Big(\frac{6.2684}{8}\Big)=-0.244$
$\text{ln}\Big(\frac{5.4115}{8}\Big)=-0.391$
$\text{ln}\Big(\frac{3.0828}{8}\Big)=-0.954$
$\text{ln}\Big(\frac{1.8899}{8}\Big)=-1.443$
$\text{ln}\Big(\frac{1.167}{8}\Big)=-1.93$
$\text{ln}\Big(\frac{0.7212}{8}\Big)=-2.406$

The half life of $^{110}Ag$ from this part of the plot is 24.4s.
Half life of $^{110}Ag = 24.4s$.

$\therefore\text{decay constant}\lambda=\frac{0.693}{24.4}=0.0284\Rightarrow\text{t}=50\text{sec,}$
The activity $\text{A = A}_{0}\text{e}^{-\lambda\text{t}}=8\times10^8\times\text{e}^{-0.0284\times50}=1.93\times10^8$

The half life period of $^{108}Ag$ from the graph is 144s.

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