5 Marks Questions - Physics STD 11 Science Questions - Vidyadip

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Question 15 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 25 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 ²¹²Bi is one hour. If 1g of pure ²¹²Bi is taken at 12.00 noon, what will be the composition of this sample at 1P.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 Bi²¹² Present = 1g

$\therefore$ at t = 1 Bi²¹² Present = 0.5g

Probability $\alpha$-decay and $\beta$-decay are in ratio $\frac{7}{13}.$

$\therefore$ Tl remained = 0.175g

$\therefore$ P₀ remained = 0.325g

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Question 35 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 = A₀

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 45 Marks

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

Time (s)	Activity (A) (10⁸ disinte- grations s^-1)	Time (s)	Activity (A) (10⁸ 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 ¹¹⁰Ag from this portion of the plot.
Use the half-life of ¹¹⁰Ag to calculate the activity corresponding to ¹⁰⁸Ag in the first 50s.
Plot In $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time for ¹⁰⁸Ag for the first 50s.
Find the half-life of ¹⁰⁸Ag.

Answer

Activities of sample containing ¹⁰⁸Ag and ¹¹⁰Ag isotopes = 8.0 × 10⁸ 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 ¹¹⁰Ag from this part of the plot is 24.4s.
Half life of ¹¹⁰Ag = 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 ¹⁰⁸Ag from the graph is 144s.

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Question 55 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 65 Marks

A point source emitting alpha particles is placed at a distance of 1m from a counter which records any alpha particle falling on its 1cm² window. If the source contains 6.0 × 10¹⁶ 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 cm² = 50000 Counts/sec.

N = N₃o of active nucleic = 6 × 10¹⁶

Total counts radiated from the source = Total surface area × 50000 counts/cm²

= 4 × 3.14 × 1 × 10⁴ × 5 × 10⁴ = 6.28 × 10⁹ $\text{Counts}=\frac{\text{dN}}{\text{dt}}$

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 75 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 85 Marks

²³⁸U decays to ²⁰⁶Pb with a half-life of 4.47 × 10⁹y. 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 2.00mg of ²³⁸U and 0.600mg of ²⁰⁶Pb. 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 U²³⁸ $=\frac{6\times10^{23}\times2\times10^{-3}}{238}=\frac{2}{238}\times10^{20}=0.05042\times10^{20}$

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 95 Marks

The half-life of ⁴⁰K is 1.30 × 10⁹y. A sample of 1.00g of pure KCI gives 160 counts/ s. Calculate the relative abundance of ⁴⁰K (fraction of ⁴⁰K 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

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 115 Marks

A vessel of volume 125cm³ 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 125 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 135 Marks

The half-life of ¹⁹⁹Au is 2.7 days.

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

Answer

198 grams of Ag contains → N₀ 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 145 Marks

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

Time t(minute):	0	25	50	75	100
Ctount rate R(10⁹s^-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 R₀ at time is t₀ = 30 × 10⁹s^-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 155 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 165 Marks

4 × 10²³ 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 175 Marks

A town has a population of 1 million. The average electric power needed per person is 300W. 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%.

Assuming 200MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day.
Assuming the fission to take place largely through ²³⁵U, at what rate will the amount of ²³⁵U decrease? Express your answer in kg per day.
Assuming that uranium enriched to 3% in ²³⁵U 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 ²³⁵U be used.

$\Rightarrow\text{x}\times\frac{3}{100}=1.264\times30$

$\Rightarrow\text{x}=1264\text{kg}$

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Question 185 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 × 10³⁰kg (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 195 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 39.9624u 39.9640u 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_value in the $\beta^-$ decay is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ca⁴⁰)]c²

= [39.9640u - 39.9626u]c²

= 0.0014 × 931MeV

= 1.3034MeV

Q_value in the $\beta^+$ decay is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ar⁴⁰) - 2m_e]c²

= [39.9640u - 39.9624u - 0.0021944u]c²

= (39.9640 - 39.9624)931MeV - 1022keV

= 1489.96keV - 1022keV

= 0.4679MeV

Q_value in the electron capture is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ar⁴⁰)]c²

= (39.9640 - 39.9624)uc²

= 1.4890 = 1.49MeV

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5 Marks Questions - Physics STD 11 Science Questions - Vidyadip