3 Marks Question - Physics STD 12 Science Questions - Vidyadip

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Question 13 Marks

Calculate the energy released by 1g of natural uranium assuming 200MeV is released in each fission event and that the fissionable isotope ²³⁵U has an abundance of 0.7% by weight in natural uranium.

Answer

1g of ‘I’ contain 0.007g U²³⁵
So, 235g contains 6.023 × 1023 atoms.
So, 0.7g contains $\frac{6.023\times10^{23}}{235}\times0.007\text{ atom}$
1 atom given 200Mev.
So, 0.7g contains $\frac{6.023\times10^{23}\times0.007\times200\times10^6\times1.6\times10^{-19}}{235}\text{J}=5.74\times10^{-8}\text{J}$

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Question 23 Marks

Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t = 0. Find the number of active nuclei at time t.

Answer

Let N₀ = No. of radioactive particle present at time t = 0
N = No. of radio active particle present at time t.
$\therefore\text{N = N}_0\text{e}^{-\lambda\text{t}}$ [$\lambda$ -Radioactive decay constant]
$\therefore$ The no.of particles decay $=\text{N}_0-\text{N}=\text{N}_0-\text{N}_0\text{e}^{-\lambda\text{t}}=\text{N}_0(1-\text{e}^{-\lambda\text{t}})$
We know, $\text{A}_0=\lambda\text{N}_0;\text{R}=\lambda\text{N}_0;\text{N}_0=\frac{\text{R}}{\lambda}$
From the above equation
$\text{N = N}_0(1-\text{e}^{-\lambda\text{t}})=\frac{\text{R}}{\lambda}(1-\text{e}^{-\lambda\text{t}})$ (substituting the value of N₀)

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Question 33 Marks

If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimise the energy?

Answer

When three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.

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Question 43 Marks

A uranium reactor develops thermal energy at a rate of 300MW. Calculate the amount of ²³⁵U being consumed every second. Average released per fission is 200MeV.

Answer

Let n atoms disintegrate per second
Total energy emitted/sec = (n × 200 × 106 × 1.6 × 10^-19)J = Power
300MW = 300 × 106 Watt = Power
300 × 106 = n × 200 × 106 × 1.6 × 10^-19
$\Rightarrow\text{n}=\frac{3}{2\times1.6}\times10^{19}=\frac{3}{3.2}\times10^{-19}$
6 × 10²³ atoms are present in 238 grams
$\frac{3}{3.2}\times10^{19}$ atoms are present in $\frac{238\times3\times10^{19}}{6\times10^{23}\times3.2}=3.7\times10^{-4}\text{g}=3.7\text{mg}.$

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Question 53 Marks

The half-life of a radioisotope is 10h. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1Ci.

Answer

$\text{t}_{\frac{1}{2}}=10\text{ hours, A}_0=1\text{ci}$
Activity after 9 hours $=\text{A}_0\text{e}^{-\lambda\text{t}}=1\times\text{e}^{\frac{-0.693}{10}\times9}=0.5359=0.539\text{ci}.$
No. of atoms left after 9th hour, $\text{A}_{9}=\lambda\text{N}_{9}$
$\Rightarrow\text{N}_9=\frac{\text{A}_9}{\lambda}=\frac{0.536\times10\times3.7\times10^{10}\times3600}{0.693}$
$=28.6176\times10^{10}\times3600=103.023\times10^{13}$
Activity after 10 hours $=\text{A}_0\text{e}^{-\lambda\text{t}}=1\times\text{e}^{\frac{-0.693}{10}\times9}=0.5\text{ci}$
No. of atoms left after 10th hour
$\text{A}_{10}=\lambda\text{N}_{10}$
$\Rightarrow\text{N}_{10}=\frac{\text{A}_{10}}{\lambda}=\frac{0.5\times3.7\times10^{10}\times3600}{\frac{0.693}{10}}$
$=26.37\times10^{10}\times3600=96.103\times10^{13}$
No.of disintegrations $=(103.023-96.103)\times10^{13}=6.92\times10^{13}$

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Question 63 Marks

Find the binding energy per nucleon of $\text{ }^{197}_{79}\text{Au}$ if its atomic mass is 196.96u.

Answer

B = (Zm_p + Nm_n - M)C²
Z = 79; N = 118; m_p = 1.007276u; M = 196.96u; m_n = 1.008665u
B = [(79 × 1.007276 + 118 × 1.008665)u - Mu]c²
= 198.597274 × 931 - 196.96 × 931 = 1524.302094
so, Binding Energy per nucleon $=\frac{1524.3}{197}=7.737.$

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Question 73 Marks

Does a nucleus lose mass when it suffers gamma decay?

Answer

Gamma rays consist of photons that are produced when a nucleus from its excited state comes to its ground state releasing energy. Since gamma rays are chargeless and massless particles, the nucleus does not suffer any loss in mass during the gamma decay.

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Question 83 Marks

A radioactive isotope is being produced at a constant rate $\frac{\text{dN}}{\text{dt}}=\text{R}$ in an experiment. The isotope has a half-life $\text{t}_{\frac{1}2{}}.$ Show that after a time $\text{t}>>\text{t}_{\frac{1}{2}},$ the number of active nuclei will become constant. Find the value of this constant.

Answer

Given: Half life period $=\text{t}_{\frac{1}{2}}$
Rate of radio active decay $=\frac{\text{dN}}{\text{dt}}=\text{R}\Rightarrow\text{R}=\frac{\text{dN}}{\text{dt}}$
Given after time $\text{t}>>\text{t}_{\frac{1}{2}},$ the number of active nuclei will become constant.
i.e. $\Big(\frac{\text{dN}}{\text{dt}}\Big)_{\text{present}}=\text{R}=\Big(\frac{\text{dN}}{\text{dt}}\Big)_{\text{decay}}$
$\therefore\text{R}=\Big(\frac{\text{dn}}{\text{dt}}\Big)_{\text{decay}}$
$\Rightarrow\text{R}=\lambda\text{N}$ [where, $\lambda$ = Radioactive decay constant, N = constant number]
$\Rightarrow\text{R}=\frac{0.693}{\text{t}_{\frac{1}{2}}}(\text{N})\Rightarrow\text{Rt}_{\frac{1}{2}}=0.693\text{N}\Rightarrow\text{N}=\frac{\text{Rt}_{\frac{1}{2}}}{0.693}$

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Question 93 Marks

A molecule. of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why?

Answer

Inside the nucleus, two protons exert nuclear force on each other. These forces are short-ranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (long-ranged). While discussing the behaviour of a hydrogen molecule, the nuclear force between the two protons is always neglected. This is because the separation between the two protons in the molecule is ~70pm which is much greater than the range of the nuclear force.

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Question 103 Marks

Find the energy liberated in the reaction:
²²³Ra → ²⁰⁹Pb + ¹⁴C.
The atomic masses needed are as follows.

²²³Ra	²⁰⁹Pb	¹⁴C
22.018u	208.981u	14.003u

Answer

²²³Ra = 223.018u; ²⁰⁹Pb = 208.981u; ¹⁴C = 14.003u.
²²³Ra → ²⁰⁹Pb + ¹⁴C
$\Delta\text{m}$ = mass ²²³Ra - mass(²⁰⁹Pb + ¹⁴C)
= 223.018 - (208.981 + 14.003) = 0.034.
Energy = $\Delta\text{M}\times\text{u}$ = 0.034 × 931 = 31.65Me.

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Question 113 Marks

Suppose we have 12 protons and 12 neutrons. We can assemble them to form either a :uMg nucleus or two 12C nuclei. In which of the two cases more energy will be liberated?

Answer

If we assemble 6 protons and 6 neutrons to form ¹²C nucleus, 92.15 MeV (product of mass number and binding energy per nucleon of carbon-12) of energy is released. Therefore, the energy released in the formation of two carbon nuclei is 184.3 MeV. On the other hand, when 12 protons and 12 neutrons are combined to form a ²⁴Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of ²⁴Mg nucleus, more energy is liberated.

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Question 123 Marks

Radioactive isotopes are produced in a nuclear physics experiment at a constant rate $\frac{\text{dN}}{\text{dt}}=\text{R}.$ An inductor of inductance 100mH, a resistor of resistance $100\Omega$ and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that $\frac{\text{i}}{\text{N}}$ remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.

Answer

$\text{R}=100\Omega;\text{ L}=100\text{mH}$
After time t, $\text{i = i}_0\Big(1-\text{e}^{\frac{-\text{t}}{\text{Lr}}}\Big)\text{ N = N}_0\big(\text{e}^{-\lambda\text{t}}\big)$
$\frac{\text{i}}{\text{N}}=\frac{\text{i}_0\big(1-\text{e}^{-\frac{\text{tR}}{\text{L}}}\big)}{\text{N}_0\text{e}^{-\lambda\text{t}}}\frac{\text{i}}{\text{N}}$ is constant i.e. independent of time.
Coefficients of t are equal $-\frac{\text{R}}{\text{L}}=-\lambda\Rightarrow\frac{\text{R}}{\text{L}}=\frac{0.693}{\text{t}_{\frac{1}{2}}}$
$=\text{t}_{\frac{1}{2}}=0.693\times10^{-3}=6.93\times10^{-4}\text{sec}.$

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Question 133 Marks

A human body excretes (removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing ⁹⁹Tc. This isotope is radioactive with a half-life of 6 hours. The activity from the body just after the injection is $6\mu\text{Ci}.$ How much time will elapse before the activity falls to $3\mu\text{Ci}?$

Answer

$\text{t}_{\frac{1}{2}}=24\text{h}$
$\therefore\text{t}_{\frac{1}{2}}=\frac{\text{t}_1\text{t}_2}{\text{t}_1+\text{t}_2}=\frac{24\times6}{24+6}=4.8\text{h}.$
$\text{A}_0=6\text{rci};\text{A}=3\text{rci}$
$\therefore\text{A}=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow3\text{rci}=\frac{6\text{rci}}{2^{\frac{\text{t}}{4.8\text{h}}}}$
$\Rightarrow\frac{\text{t}}{24.8\text{h}}=2\Rightarrow\text{t}=4.8\text{h}$

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Question 143 Marks

Assume that the mass of a nucleus is approximately given by M = Am_p where A is the mass number. Estimate the density of matter in kg/m³ inside a nucleus. What is the specific gravity of nuclear matter?

Answer

$\text{M = Am}_{\text{p}},\text{f}=\frac{\text{M}}{\text{V}},\text{m}_{\text{p}}=1.007276\text{u}$
$\text{R = R}_0\text{A}^{\frac{1}{3}}=1.1\times10^{-15}\text{A}^{\frac{1}{3}},\\\text{u}=1.6605402\times10^{-27}\text{kg}$
$=\frac{\text{A}\times1.007276\times1.6605402\times10^{-27}}{\frac{4}{3}\times3.14\times\text{R}^3}$
$=0.300159\times10^{18}=3\times10^{17}\text{kg/m}^3.$
‘f’ in CGS = Specific gravity $=3\times10^{14}.$

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Question 153 Marks

How much energy is released in the following reaction?
$\text{ }^7\text{Li + p}\rightarrow\alpha+\alpha.$
Atomic mass of ⁷Li = 7.0160u and that of ⁴He = 4.0026u.

Answer

$\text{Li}^7+\text{p}\rightarrow\text{l}+\alpha+\text{E};\text{Li}^7=7.016\text{u}$
$\alpha=\text{ }^4\text{He}=4.0026\text{u};\text{p}=1.007276\text{u}$
$\text{E}=\text{Li}^7+\text{P}-2\alpha=(7.016+1.007276)\text{u}\\-(2\times4.0026)\text{u}=0.018076\text{u}.$
$\Rightarrow0.018076\times931=16.828=16.83\text{MeV}.$

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Question 163 Marks

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
$\text{ }^{12}\text{N}\rightarrow{ }^{12}\text{C}^*+\text{e}^++\text{v}$
$\text{ }^{12}\text{C}^*\rightarrow\text{ }^{12}\text{C}+\gamma(4.43\text{MeV}).$
The atomic mass of ¹²N is 12.018613u.

Answer

Given:
Atomic mass of ¹²N, m(¹²N) = 12.018613u
¹²N → ¹²C* + e⁺ + v
¹²C* → ¹²C + Y (4.43MeV)
Net reaction is given by
¹²N → ¹²C + e⁺ + v + Y (4.43MeV)
Q_value of the $\beta^+$ decay will be
Q_value= [m(¹²N) - (m(¹²C*) + 2m_e)]c²
= [12.018613 × 931MeV - (12 × 931 + 4.43) MeV - (2 × 511)keV]
= [11189.3287 - 11176.43 - 1.022]MeV
= 11.8767MeV = 11.88MeV
The maximum kinetic energy of beta particle will be 11.88MeV, assuming that neutrinos have zero energy.

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Question 173 Marks

A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes.

What is the decay constant of the sample?
What is its half-life?

Answer

$\text{A = 200, A}_0 = 500, \text{t = 50 min}$

$\text{A = A}_0\text{e}^{-\lambda\text{t}}$

$200=500\times\text{e}^{-50\times60\times\lambda}$

$\Rightarrow\lambda=3.05\times10^{-4}\text{s}.$

$\text{t}_{\frac{1}{2}}=\frac{0693}{\lambda}=\frac{0.693}{0.000305}=2272.13\sec=38\text{min}$

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Question 183 Marks

A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life $\tau.$ Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

Answer

$\text{Q = qe}^{\frac{-\text{t}}{\text{CR}}};\text{A = A}_0\text{e}^{-\lambda\text{t}}$
$\frac{\text{Energy}}{\text{Activity}}=\frac{1\text{q}^2\times\text{e}^{\frac{-2\text{t}}{\text{CR}}}}{2\text{CA}_0\text{e}^{-\lambda\text{t}}}$
Since the term is independent of time, so their coefficients can be equated,
So, $\frac{2\text{t}}{\text{CR}}=\lambda\text{t}$
$\lambda=\frac{2}{\text{CR}}$
$\frac{1}{\tau}=\frac{2}{\text{CR}}$
$\text{R}=2\frac{\tau}{\text{C}}$

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Question 193 Marks

²²⁸Th emits an alpha particle to reduce to ²²⁴Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
$\text{ }^{228}\text{Th}\rightarrow\text{ }^{224}\text{Ra}^*+\alpha$
$\text{ }^{224}\text{Ra}^*\rightarrow\text{ }^{224}\text{Ra}+\gamma(217\text{kev}).$
Atomic mass of ²²⁸Th is 228.028726u, that of ²²⁴Ra is 224.020196u and that of $\text{ }^4_2\text{He}$ is 4.00260u.

Answer

Mass $\text{ }^{228}\text{Th}=228.028726\text{u};\text{ }^{224}\text{Ra}=224.020196\text{u}$
$\alpha=\text{ }^4_2\text{He}\rightarrow4.00260\text{u}$
$\text{ }^{228}\text{Th}\rightarrow\text{ }^{224}\text{Ra}^*+\alpha$
$\text{ }^{224}\text{Ra}^*\rightarrow\text{ }^{224}\text{Ra + v}(217\text{kev})$
Now, Mass of $\text{ }^{224}\text{Ra}^* = 224.020196 \times 931 + 0.217\text{ Mev} $
$= 208563.0195\text{Mev.}$
KE of $\alpha=\text{E}^{226}\text{Th}-\text{E}(\text{ }^{224}\text{Ra}^*+\alpha)$
$= 228.028726\times 931-[208563.0195 + 4.00260\times931]$
$= 5.30383\text{Mev}= 5.304\text{Mev.}$

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Question 203 Marks

Carbon (Z = 6) with mass number 11 decays to boron (Z = 5).

Is it a $\beta^+-\text{decay}$ or a $\beta^--\text{decay}?$
The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?

Answer

$\text{P}\rightarrow\text{n + e}^++\text{v}$ Hence it is a $\beta^+\text{decay}.$
Let the total no. of atoms be 100 N₀.

	Carbon	Boron
Initially	90N₀	10N₀
Finally	10N₀	90N₀

Now, $10\text{N}_0=90\text{N}_0\text{e}^{-\lambda\text{t}}\Rightarrow\frac{1}9{}=\text{e}^{\frac{-0.693}{20.3}\times\text{t}}$ $\Big[$because $\text{t}_{\frac{1}2{}}=20.3\text{min}\Big]$
$\Rightarrow\text{In}\frac{1}9{}=\frac{-0.693}{20.3}\text{t}\Rightarrow\text{t}=\frac{2.1972\times20.3}{0.693}=64.36=64\text{min}.$

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Question 213 Marks

In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Greater linear momentum?

Answer

Two photons having equal liner momentum have equal wavelengths as here for both the photons the direction and magnitude of linear momentum will be same. For the rest of the options, magnitude will be same but nothing can be said about the direction of the photons.
Hence the correct option is D.

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Question 223 Marks

Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5kT equals the Coulomb potential energy at 2fm.

Answer

$\text{PE}=\frac{\text{Kq}_1\text{q}_2}{\text{r}}=\frac{9\times10^9\times(2\times1.6\times10^{-19})^2}{\text{r}} \ ...(1)$
$1.5\text{KT}=1.5\times1.38\times10^{-23}\times\text{T} \ ...(2)$
Equating (1) and (2) $1.5\times1.38\times10^{-23}\times\text{T}=\frac{9\times10^9\times10.24\times10^{-38}}{2\times10^{-15}}$
$\Rightarrow\text{T}=\frac{9\times10.24\times10^{-38}}{2\times10^{-15}\times1.5\times1.38\times10^{-23}}$
$=22.26087\times10^9\text{K}=2.23\times10^{10}\text{K}$

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Question 233 Marks

Calculate the Q-value of the fusion reaction
⁴He + ⁴He = ⁸Be.
Is such a fusion energetically favourable? Atomic mass of ⁸Be is 8.0053u and that of ⁴He is 4.0026u.

Answer

⁴H + ⁴H → ⁸Be
M(²H) → 4.0026u
M(⁸Be) → 8.0053u
Q value = [2M(²H) - M(⁸Be)] = (2 × 4.0026 - 8.0053)u
= -0.0001u = -0.0931Mev = -93.1Kev.

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Question 243 Marks

Calculate the energy that can be obtained from 1kg of water through the fusion reaction
²H + ²H → ³H + p.
Assume that 1.5 × 10^-2% of natural water is heavy water D₂O (by number of molecules) and all the deuterium is used for fusion.

Answer

Given:
18g of water contains 6.023 × 10²³molecules.
$\therefore1000\text{g}$ of water $=\frac{6.023\times10^{23}\times1000}{18}=3.346\times10^{25}$ molecules
% of deuterium $=3.346\times10^{25}\times\frac{0.015}{100}=0.05019\times10^{23}$
Energy of deuterium $=30.4486\times10^{25}$
= [2 × m(²H) - m(³H) - m_p]c²
= (2 × 2.014102u - 3.016049u - 1.007276u)c²
= 0.004879 × 931Me
= 4.542349Me
= 7.262 × 10^-13J
Total energy = 0.05019 × 10²³ × 7.262 × 10^-13J
= 3644MJ

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Question 253 Marks

The decay constant of ²³⁸U is 4.9 × 10^-18 S^-1.

What is the average-life of ²³⁸U?
What is the half-life of ²³⁸U?
By what factor does the activity of a ²³⁸U sample decrease in 9 × 10⁹ years?

Answer

$\lambda=4.9\times10^{-18}\text{s}^{-1}$

Avg. life of $\text{ }^{238}\text{U}=\frac{1}{\lambda}=\frac{1}{4.9\times10^{-18}}=\frac{1}{4.9}\times10^{-18}\sec.$

$=6.47\times10^{3}\text{years}.$

Half life of uranium $=\frac{0.693}{\lambda}=\frac{0.693}{4.9\times10^{-18}}=4.5\times10^9\text{years}.$
$\text{A}=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow\frac{\text{A}_0}{\text{A}}=2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=2^2=4.$

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Question 263 Marks

When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ¹⁴C per gram per minute. A sample from an ancient piece of charcoal shows ¹⁴C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of ¹⁴C is 5730y.

Answer

$\text{A}_0=15.3;\text{ A}=12.3;\text{t}_{\frac{1}{2}}=5730\text{ year}$
$\lambda=\frac{0.6931}{\text{T}_{\frac{1}{2}}}=\frac{0.6931}{5730}\text{yr}^{-1}$
Let the time passed be t,
We know $\text{A = A}_0\text{e}^{-\lambda\text{t}}-\frac{0.6931}{5730}\times\text{t}$
$\Rightarrow12.3=15.3\times\text{e}$
$\Rightarrow\text{t}=1804.3\text{ years.}$

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Question 273 Marks

Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs?

Answer

Neutrons are chargeless particles and they exert only short range nuclear forces on each other. If we have two pairs of neutrons and the separation between them is same in both the pairs. The force between the neutrons will be of same magnitude for the two pairs until there is some other influence on any of them.

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Question 283 Marks

Is it easier to take out a nucleon from carbon or from iron? Fi-om iron or from lead?

Answer

Binding energy per nucleon of a nucleus is defined as the energy required to break-off a nucleon from it.

As the binding energy per nucleon of iron is more than that of carbon, it is easier to take out a nucleon from carbon than iron.
As the binding energy per nucleon of iron is more than that of lead. Therefore, it is easier to take out a nucleon from lead as compared to iron.

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Question 293 Marks

The count rate from a radioactive sample falls from 4.0 × 10⁶ per second to 1.0 × 10⁶per second in 20 hours. What will be the count rate 100 hours after the beginning?

Answer

$\text{A}_0=4\times10^5$ disintegration/sec

$\text{A}'=1\times10^6$ dis/sec; t = 20 hours.

$\text{A}'=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=\frac{\text{A}_0}{\text{A}'}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=4$

$\Rightarrow\frac{\text{t}}{\text{t}_{\frac{1}{2}}}=2\Rightarrow\text{t}^{\frac{1}{2}}=\frac{\text{t}}{2}=\frac{20\text{ hours}}{2}=10\text{ hours}.$

$\text{A}''=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow\text{A}''=\frac{4\times10^6}{2^{\frac{100}{10}}}$

$=0.00390625\times10^6=3.9\times10^3$ dintegrations/sec.

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Question 303 Marks

$\text{ }^{197}_{80}\text{Hg}$ decay to $\text{ }^{197}_{79}\text{Au}$ through electron capture with a decay constant of 0.257 per day.

What other particle or particles are emitted in the decay?
Assume that the electron is captured from the K shell. Use Moseley's law $\sqrt{\text{v}}=\text{a(Z}-\text{b})$ with $\text{a}=4.95\times10^7\text{s}^{-\frac{1}{2}}$ and b = 1 to find the wavelength of the $\text{K}_{\alpha}$ X-ray emitted following the electron capture.

Answer

$\text{P + e}\rightarrow\text{n + v}$ neutrino $\big[\text{a}\rightarrow4.95\times10^7\text{s}^{-\frac{1}{2}};\text{b}\rightarrow1\big]$
$\sqrt{\text{f}}=\text{a(z}-\text{b})$

$\Rightarrow\sqrt{\frac{\text{c}}{\lambda}}=4.95\times10^7(79-1)=4.95\times10^7\times78$

$\Rightarrow\frac{\text{c}}{\lambda}=(4.95\times78)^2\times10^{14}$

$\Rightarrow\lambda=\frac{3\times10^8}{14903.2\times10^{14}}$

$=2\times10^{-5}\times10^{-6}=2\times10^{-4}\text{m}=20\text{pm}$

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3 Marks Question - Physics STD 12 Science Questions - Vidyadip