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Class 12 Physics: Atoms and Nuclei Chapter Notes and Important Questions
GIST
| Topic | Details |
|---|---|
| Thomson’s model of atom | Every atom consists of a positively charged sphere in which electrons are embedded like seeds in a watermelon. Drawbacks: it could not explain large-angle scattering and the origin of spectral series. |
| Rutherford’s model of atom |
|
| Distance of closest approach of the alpha particle in the α-particle scattering experiment | \(r_0=\dfrac{2kZe^2}{\frac12 mv^2}\) |
| Impact parameter of the alpha particle | \(b=\dfrac{kZe^2\cot\frac{\theta}{2}}{\frac12 mv^2}\) |
| Bohr’s model of atom | Limitation: applicable only to hydrogen-like atoms; it could not explain the splitting of spectral lines (it does not consider the electrostatic force among the electrons). |
| Orbit radius and speed of the electron around the nucleus | \(r=\dfrac{n^2h^2}{4\pi^2 mke^2}\), \(\ v=\dfrac{2\pi ke^2}{nh}\) |
| Energy of the electron in the nth orbit of the hydrogen atom | \(E_n=-\dfrac{2\pi^2mk^2e^4}{n^2h^2}=-\dfrac{13.6}{n^2}\ \text{eV}=-\dfrac{2.18\times10^{-18}}{n^2}\ \text{joule}\) |
| Wave number | \(\dfrac1\lambda=R\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]\) |
UNIT 8: ATOMS AND NUCLEI – QUESTION BANK
One Mark Questions
In Bohr’s model of the hydrogen atom, name the physical quantity which equals an integral multiple of \(\dfrac{h}{2\pi}\).
Answer: Angular momentum of the electron, \(L=mvr=\dfrac{nh}{2\pi}\).
What is the relation between 'n' and the radius 'r' of the orbit of the electron in a hydrogen atom according to Bohr’s theory?
Answer: \(r\propto n^2\).
What is Bohr’s quantisation condition?
Answer: The electron revolves only in those orbits for which its angular momentum is an integral multiple of \(\dfrac{h}{2\pi}\): \(mvr=\dfrac{nh}{2\pi}\), \(n=1,2,3,\ldots\)
How is a β particle different from an electron?
Answer: A β-particle is an electron (or positron), but it is created inside the nucleus when a neutron changes to a proton and is then emitted with high speed. An ordinary electron is a constituent of the atom outside the nucleus.
Draw a graph between the number of nuclei undecayed and time for a radioactive substance.
Answer: The number of undecayed nuclei falls exponentially with time, \(N=N_0e^{-\lambda t}\).
Among the alpha, beta and gamma radiations, which are the radiations affected by a magnetic field?
Answer: Alpha and beta radiations (they are charged particles). Gamma rays are not affected.
Why do α particles have high ionising power?
Answer: An α-particle has a large mass and a charge of \(+2e\) and moves slowly compared with β and γ radiation, so it strongly attracts electrons of the atoms it passes and ionises many of them.
Write the relationship between the half-life and the average life of a radioactive substance.
Answer: \(\tau=\dfrac{T_{1/2}}{0.693}=1.44\,T_{1/2}\).
Why is heavy water often used in a nuclear reactor as a moderator?
Answer: Heavy water slows down fast neutrons through elastic collisions with deuterium nuclei and absorbs very few neutrons, so the neutrons are available for further fission.
Why is a neutron very effective as a bombarding particle in a nuclear reaction?
Answer: A neutron has no charge, so it is not repelled by the positively charged nucleus and can approach and enter the nucleus even with low energy.
Why are the control rods made of cadmium?
Answer: Cadmium absorbs neutrons very strongly, so the rods can control the rate of the chain reaction.
Name the phenomenon by which energy is produced in stars.
Answer: Nuclear fusion.
Define atomic mass unit. Write its energy equivalent in MeV.
Answer: One atomic mass unit (u) is \(\dfrac{1}{12}\) of the mass of one atom of carbon-12. \(1\ \text{u}=931.5\ \text{MeV}\).
What was the drawback of Rutherford’s model of the atom?
Answer: It could not explain the stability of the atom (an orbiting electron should radiate energy and spiral into the nucleus) or the line spectra of atoms.
Name the series of the hydrogen spectrum which has the least wavelength.
Answer: Lyman series.
Any two protons repel each other, so how is it possible for them to remain together in a nucleus?
Answer: The nuclear force between two protons is about 100 times stronger than the electrostatic repulsion between them.
Define radioactive decay constant.
Answer: The decay constant \(\lambda\) is the ratio of the rate of decay to the number of undecayed nuclei, \(\lambda=\dfrac{-dN/dt}{N}\). It is the reciprocal of the time in which the number of nuclei falls to \(\dfrac1e\) of its initial value.
Write two important inferences drawn from the Geiger–Marsden experiment.
Answer:
- Most of the atom is empty space, since most α-particles pass undeflected.
- The entire positive charge and almost all the mass of the atom are concentrated in a very small region at the centre, the nucleus (size about \(10^{-15}\) m).
What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus (\(Z=79\))?
Answer: At the closest approach all the kinetic energy becomes potential energy: \(K=\dfrac{kZe\cdot e}{r_0}\)
\(r_0=\dfrac{9\times10^9\times79\times(1.6\times10^{-19})^2}{5\times1.6\times10^{-13}}\approx2.3\times10^{-14}\ \text{m}\).
Two nuclei have mass numbers in the ratio 1 : 8. What is the ratio of their nuclear radii?
Answer: \(R\propto A^{1/3}\), so \(\dfrac{R_1}{R_2}=\left(\dfrac18\right)^{1/3}=\dfrac12\), that is 1 : 2.
What is the relation between the radius of the nucleus and the mass number?
Answer: \(R=R_0A^{1/3}\), i.e. \(R\propto A^{1/3}\).
What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1 : 4?
Answer: 1 : 1. Nuclear density is independent of the mass number.
Two Marks Questions
Define the activity of a given radioactive substance. Write its SI unit.
Answer: Activity is the number of nuclei disintegrating per second, \(R=-\dfrac{dN}{dt}=\lambda N\). Its SI unit is the becquerel (Bq), 1 Bq = 1 decay per second.
(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents, the neutrons and protons. Explain.
Answer: When nucleons combine to form a nucleus, energy equal to the binding energy is released. By \(E=\Delta m\,c^2\), this energy comes from a loss of mass, called the mass defect, so the nucleus has less mass than its separate nucleons.
The radioactive isotope D decays according to the sequence \(D\xrightarrow{\ \beta\text{-particle}\ }D_1\xrightarrow{\ \alpha\text{-particle}\ }D_2\). If the mass number and atomic number of \(D_2\) are 176 and 71 respectively, what are (i) the mass number and (ii) the atomic number of D?
Answer: An α-particle removes 4 from mass number and 2 from atomic number, so \(D_1\) has \(A=180\), \(Z=73\). A β− particle raises \(Z\) by 1 without changing \(A\), so \(D\) has \(A=180\), \(Z=72\).
(i) Mass number of D = 180. (ii) Atomic number of D = 72.
Why is it found experimentally difficult to detect neutrinos in nuclear β-decay?
Answer: Neutrinos have no charge and almost zero mass, and they interact extremely weakly with matter, so they pass through detectors almost always without any interaction.
Draw a plot of the potential energy of a pair of nucleons as a function of their separation. What is the significance of negative potential energy in the graph drawn?
Answer: The potential energy is negative for separations greater than about \(0.8\) fm, which shows that the nuclear force is attractive there and the nucleons are bound; energy has to be supplied to separate them.
If 70% of a given radioactive sample is left undecayed after 20 days, what percentage of the original sample will get decayed in 60 days?
Answer: 60 days is three intervals of 20 days. Each interval leaves 70% of what was present, so the fraction left is \(0.7^3=0.343\), i.e. 34.3%.
Percentage decayed \(=100-34.3=65.7\%\).
How is the neutron to proton ratio affected during (i) β decay (ii) α decay?
Answer: (i) In β− decay a neutron changes into a proton, so the ratio \(N/Z\) decreases.
(ii) In α decay two protons and two neutrons are lost; for a heavy nucleus \(N>Z\), so \(\dfrac{N-2}{Z-2}\) is larger, i.e. the ratio increases.
A radioactive sample having N nuclei has activity R. Write an expression for its half-life in terms of R and N.
Answer: \(R=N\lambda\) and \(T_{1/2}=\dfrac{0.693}{\lambda}\), so \(T_{1/2}=\dfrac{0.693\,N}{R}\).
Tritium has a half-life of 12.5 years against beta decay. What fraction of a sample of pure tritium will remain undecayed after 25 years?
Answer: 25 years = 2 half-lives, so \(\dfrac{N}{N_0}=\left(\dfrac12\right)^2=\dfrac14\). One-fourth of the sample remains.
What percentage of a given mass of a radioactive substance will be left undecayed after 5 half-life periods?
Answer: \(\dfrac{N}{N_0}=\dfrac{1}{2^n}=\dfrac1{32}=3.125\%\).
A radioactive nucleus A decays as \(A\xrightarrow{\ \beta\ }A_1\xrightarrow{\ \gamma\ }A_2\). If the mass number and atomic number of \(A_1\) are 180 and 73 respectively, find the mass number and atomic number of A and \(A_2\).
Answer: A γ-ray changes neither \(A\) nor \(Z\), so \(A_2\) has \(A=180\), \(Z=73\). A β− decay raises \(Z\) by 1, so A has \(A=180\), \(Z=72\).
Two nuclei P and Q have an equal number of atoms at \(t=0\). Their half-lives are 3 and 9 hours respectively. Compare the rates of disintegration after 18 hours from the start.
Answer: After 18 h, P has gone through 6 half-lives, so \(N_P=\dfrac{N_0}{64}\); Q has gone through 2 half-lives, so \(N_Q=\dfrac{N_0}{4}\). With \(R=\lambda N\) and \(\lambda\propto\dfrac1{T_{1/2}}\):
\(\dfrac{R_P}{R_Q}=\dfrac{(1/3)(1/64)}{(1/9)(1/4)}=\dfrac{3}{16}\).
The ratio is 3 : 16.
What is the neutron multiplication factor? For what value of this is a nuclear reactor said to be critical?
Answer: The neutron multiplication factor \(K\) is the ratio of the number of fission neutrons in one generation to the number in the preceding generation. The reactor is critical when \(K=1\) (steady chain reaction).
Four nuclei of an element fuse together to form a heavier nucleus. If the process is accompanied by release of energy, which of the two, the parent or the daughter nuclei, would have higher binding energy per nucleon? Justify your answer.
Answer: The daughter nucleus. Energy is released only when the product is more tightly bound, i.e. its total binding energy exceeds that of the original nuclei; the energy released equals the gain in binding energy. A more tightly bound system has a higher binding energy per nucleon.
Find the ratio of the energies of photons produced due to transition of an electron of a hydrogen atom from (i) its second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
Answer: (i) \(E_2\to E_1=13.6\left(1-\dfrac14\right)=10.2\ \text{eV}\)
(ii) \(E_\infty\to E_1=13.6\left(1-0\right)=13.6\ \text{eV}\)
Ratio \(=\dfrac{10.2}{13.6}=\dfrac34\), i.e. 3 : 4.
The ground state energy of the hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer: Total energy \(E=-13.6\) eV. Kinetic energy \(K=-E=+13.6\) eV. Potential energy \(U=2E=-27.2\) eV.
Using Bohr’s postulates of the atomic model, derive the expression for the radius of the nth electron orbit. Hence obtain the expression for the Bohr radius.
Answer: For an electron in the nth orbit the Coulomb force gives the centripetal force:
\[\frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}\]
By Bohr’s quantisation condition, \(mvr=\dfrac{nh}{2\pi}\), so \(v=\dfrac{nh}{2\pi mr}\). Substituting in the first equation:
\[r_n=\frac{n^2h^2\varepsilon_0}{\pi me^2}\]
For \(n=1\) this gives the Bohr radius \(a_0=\dfrac{h^2\varepsilon_0}{\pi me^2}\approx0.53\ \text{Å}\), so \(r_n=n^2a_0\).
Using the Rutherford model of the atom, derive the expression for the total energy of the electron in the hydrogen atom. What is the significance of the total negative energy possessed by the electron?
Answer: The Coulomb force supplies the centripetal force: \(\dfrac{mv^2}{r}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}\), so the kinetic energy is
\[K=\tfrac12mv^2=\frac{e^2}{8\pi\varepsilon_0r}\]
The potential energy is \(U=-\dfrac{e^2}{4\pi\varepsilon_0r}\). The total energy is
\[E=K+U=-\frac{e^2}{8\pi\varepsilon_0r}\]
The negative sign shows that the electron is bound to the nucleus; energy equal to \(|E|\) must be supplied to free it from the atom.
Three Marks Questions
Obtain the binding energy of the nuclei \(^{56}_{26}\text{Fe}\) and \(^{209}_{83}\text{Bi}\) in MeV from the following data: \(m_H=1.007825\) amu, \(m_n=1.008665\) amu, \(m(^{56}_{26}\text{Fe})=55.934939\) amu, \(m(^{209}_{83}\text{Bi})=208.980388\) amu, 1 amu = 931.5 MeV.
Answer: Fe-56 (26 protons, 30 neutrons):
\(\Delta m=26(1.007825)+30(1.008665)-55.934939=0.528461\) amu
\(BE=0.528461\times931.5\approx492.3\) MeV, i.e. about 8.79 MeV per nucleon.
Bi-209 (83 protons, 126 neutrons):
\(\Delta m=83(1.007825)+126(1.008665)-208.980388=1.760877\) amu
\(BE=1.760877\times931.5\approx1640.3\) MeV, i.e. about 7.85 MeV per nucleon.
Plot a graph of the binding energy per nucleon as a function of mass number for a large number of nuclei. Explain the energy release in the process of nuclear fission from the above plot. Write a typical nuclear reaction in which a large amount of energy is released in the process of nuclear fission.
Answer: The binding energy per nucleon is about 7.6 MeV for heavy nuclei (\(A\approx240\)) and about 8.5 MeV for medium nuclei (\(A\approx120\)). When a heavy nucleus splits into two medium nuclei, the products are more tightly bound, so the gain of about 0.9 MeV per nucleon, \(\approx0.9\times240\approx200\) MeV in all, is released as energy. A typical fission reaction:
\[^{235}_{92}\text{U}+^{1}_{0}n\to\,^{141}_{56}\text{Ba}+^{92}_{36}\text{Kr}+3\,^{1}_{0}n+\text{energy (about 200 MeV)}\]
Define the activity of a radionuclide. Write its SI unit. Plot a graph of activity of a radioactive species versus time. How long will a radioactive isotope, whose half-life is T years, take for its activity to reduce to 1/8th of its initial value?
Answer: Activity is the rate of disintegration, \(R=-\dfrac{dN}{dt}=\lambda N\); its SI unit is the becquerel (Bq). Activity falls exponentially with time:
\(\dfrac{R}{R_0}=\dfrac18=\left(\dfrac12\right)^3\), so 3 half-lives are needed: the time is \(3T\) years.
Plot a graph of the potential energy of a pair of nucleons as a function of their separation. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.
Answer: (i) The force is attractive for separations greater than about 0.8 fm (the potential energy rises with distance after the minimum). (ii) It is repulsive for separations less than about 0.8 fm (the potential energy rises steeply as the nucleons come closer).
Features:
- Nuclear forces are very strong, short-range forces (negligible beyond a few fm).
- They are charge independent: the force between two protons, two neutrons, or a proton and a neutron is nearly the same.
Plot a graph of the potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
Answer:
- The force is attractive and much stronger than the electrostatic force at distances of the order of 1 fm, and it falls to nearly zero beyond a few fm, so it is short ranged.
- It becomes repulsive at distances less than about 0.8 fm, so the nucleons cannot collapse into each other.
Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release of energy in the processes of nuclear fission and nuclear fusion can be explained.
Answer: The curve rises steeply for light nuclei, has its maximum (about 8.8 MeV per nucleon) near \(A=56\), and then falls slowly for heavy nuclei.
- Fission: a heavy nucleus (\(A\approx240\), about 7.6 MeV per nucleon) splits into two medium nuclei (\(A\approx120\), about 8.5 MeV per nucleon). The products are more tightly bound, so energy is released.
- Fusion: two light nuclei (for example \(A\approx2\), with low binding energy per nucleon) join to form a heavier nucleus with a higher binding energy per nucleon, so energy is released.
Two radioactive materials X1 and X2 have decay constants \(10\lambda\) and \(\lambda\) respectively. If initially they have the same number of nuclei, find the time after which the ratio of the nuclei of X1 to that of X2 will be \(1/e\).
Answer: \(N=N_0e^{-\lambda t}\), so \(\dfrac{N_1}{N_2}=\dfrac{e^{-10\lambda t}}{e^{-\lambda t}}=e^{-9\lambda t}=\dfrac1e\).
Hence \(9\lambda t=1\), \(t=\dfrac{1}{9\lambda}\).
At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%. Calculate (i) the mean life of the nuclei, (ii) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.
Answer: The number halves in 10 s, so \(T_{1/2}=10\) s.
(i) \(\lambda=\dfrac{0.693}{10}=0.0693\ \text{s}^{-1}\); mean life \(\tau=\dfrac1\lambda=14.43\) s.
(ii) \(\dfrac{N}{N_0}=\dfrac{6.25}{100}=\dfrac1{16}=\left(\dfrac12\right)^4\), so 4 half-lives, \(t=4\times10=40\) s.
(a) Draw the energy level diagram showing the emission of β-particles followed by γ-rays by a \(^{60}_{27}\text{Co}\) nucleus. (b) Plot the distribution of kinetic energy of β− particles and state why the energy spectrum is continuous.
Answer: (a) \(^{60}_{27}\text{Co}\) decays by β− emission to an excited level of \(^{60}_{28}\text{Ni}\), which de-excites to the ground state by emitting two γ-rays of 1.17 MeV and 1.33 MeV.
(b) The spectrum is continuous because the energy released in each decay is shared between the β-particle and an antineutrino in all possible proportions, so the β-particle can have any kinetic energy from zero up to a maximum \(K_{max}\).
Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when an electron in the hydrogen atom undergoes a transition from a higher energy state (quantum number \(n_i\)) to a lower state (\(n_f\)). When the electron in the hydrogen atom jumps from energy state \(n_i=4\) to \(n_f=3,2,1\), identify the spectral series to which the emission lines belong.
Answer: By the Bohr frequency condition, \(h\nu=E_i-E_f\). With \(E_n=-\dfrac{me^4}{8\varepsilon_0^2h^2n^2}\):
\[\nu=\frac{E_i-E_f}{h}=\frac{me^4}{8\varepsilon_0^2h^3}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\]
Transitions from \(n_i=4\): to \(n_f=3\) belongs to the Paschen series; to \(n_f=2\) the Balmer series; to \(n_f=1\) the Lyman series.
Given the value of the ground state energy of the hydrogen atom as −13.6 eV, find out its kinetic and potential energy in the ground and second excited states.
Answer: For an electron in a Bohr orbit, \(K=-E\) and \(U=2E\).
Ground state (\(n=1\)): \(E=-13.6\) eV, \(K=+13.6\) eV, \(U=-27.2\) eV.
Second excited state (\(n=3\)): \(E=-\dfrac{13.6}{9}=-1.51\) eV, \(K=+1.51\) eV, \(U=-3.02\) eV.
State the law of radioactive decay. If \(N_0\) is the number of radioactive nuclei in the sample at some initial time \(t_0\), find out the relation to determine the number \(N\) present at a subsequent time. Plot a graph of \(N\) as a function of time.
Answer: Law: the rate of decay of a radioactive sample at any instant is directly proportional to the number of undecayed nuclei present at that instant: \(\dfrac{dN}{dt}=-\lambda N\). Integrating from \(t_0\) (when \(N=N_0\)) to \(t\):
\[\int_{N_0}^{N}\frac{dN}{N}=-\lambda\int_{t_0}^{t}dt\ \Rightarrow\ N=N_0e^{-\lambda(t-t_0)}\]
Distinguish between isotopes and isobars. Give one example for each of the species. A radioactive isotope has a half-life of 5 years. How long will it take the activity to reduce to 3.125%?
Answer: Isotopes are nuclides with the same atomic number but different mass numbers, for example \(^{1}_{1}\text{H}\) and \(^{2}_{1}\text{H}\). Isobars are nuclides with the same mass number but different atomic numbers, for example \(^{14}_{6}\text{C}\) and \(^{14}_{7}\text{N}\).
\(3.125\%=\dfrac1{32}=\left(\dfrac12\right)^5\), so 5 half-lives are needed: \(t=5\times5=25\) years.
(a) Write symbolically the β− decay process of \(^{32}_{15}\text{P}\). (b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life.
Answer: (a) \(^{32}_{15}\text{P}\to\,^{32}_{16}\text{S}+e^-+\bar\nu\)
(b) The number of nuclei that decay between \(t\) and \(t+dt\) is \(\lambda N\,dt=\lambda N_0e^{-\lambda t}dt\), and each has lived for time \(t\). Average life:
\[\tau=\frac{1}{N_0}\int_0^\infty t\,\lambda N_0e^{-\lambda t}\,dt=\frac1\lambda\]
Since \(T_{1/2}=\dfrac{0.693}{\lambda}\), we get \(\tau=\dfrac{T_{1/2}}{0.693}=1.44\,T_{1/2}\).
The energy levels of an atom are as shown below. (a) Which of them will result in the transition of a photon of wavelength 275 nm? (b) Which transition corresponds to the emission of radiation of maximum wavelength?
Answer: (a) \(E=\dfrac{hc}{\lambda}=\dfrac{1240\ \text{eV nm}}{275\ \text{nm}}=4.5\) eV, which equals the gap of transition B (as given in the source key).
(b) \(\lambda\propto\dfrac1E\), so the maximum wavelength corresponds to the smallest energy gap, which is transition A (0 to \(-2\) eV, 2 eV).
Value Based Questions
(1) A farmer in a village was worried about the poor yield of the soil. Rohan, during his visit to his native place, happened to meet this farmer and suggested that he use a phosphate fertilizer incorporated with radio phosphorus. He also explained that phosphorus will be taken by the plant for its growth and radio phosphorus will increase the yield. The farmer thanked him for his valuable suggestion. (a) Suggest the moral value that you derive from Rohan. (b) A radioactive isotope has a half-life of T years. How long will it take for the activity to reduce to (i) 3.125% (ii) 1% of its original value?
Answer: (a) Concern for society and the locality, awareness, presence of mind.
(b) (i) \(3.125\%=\dfrac1{32}=\left(\dfrac12\right)^5\), so \(t=5T\).
(ii) \(\dfrac{N}{N_0}=0.01=e^{-\lambda t}\), so \(t=\dfrac{\ln100}{\lambda}=\dfrac{4.605\,T}{0.693}\approx6.64\,T\).
(2) Akil and Nikhil are arguing about the estimation of the age of a specimen by a scientific method. Akil said that there is no way of finding the age of a specimen scientifically. But Nikhil argued that there should be one method, but he is not aware of it. Tarun, who is witnessing this argument, convinced them not to proceed with the argument. He said that the age of the specimen can be estimated by noting the drop in the activity of carbon C-14 after the organism is dead. Listening to the explanation, both of them were convinced and also felt happy as they had learnt a new concept. (a) What moral value do you observe in Tarun? (b) Obtain the amount of Co-60 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co-60 is 5.3 years.
Answer: (a) Readiness to teach his juniors; concern for his juniors learning.
(b) \(R=8.0\ \text{mCi}=8.0\times10^{-3}\times3.7\times10^{10}=2.96\times10^{8}\) Bq
\(\lambda=\dfrac{0.693}{5.3\times3.156\times10^{7}\ \text{s}}\approx4.14\times10^{-9}\ \text{s}^{-1}\)
\(N=\dfrac R\lambda=\dfrac{2.96\times10^8}{4.14\times10^{-9}}\approx7.1\times10^{16}\) nuclei
Mass \(=\dfrac{7.1\times10^{16}}{6.022\times10^{23}}\times60\ \text{g}\approx7.1\times10^{-6}\) g.
(3) (a) Shyam saw his younger brother wondering about a question which deals with the emission of light from a vapour lamp. He was anxious to know how different colours were being emitted by different lamps. He also saw mercury and sodium vapour lamps in the physics lab and was curious to know what is inside the lamps. On seeing his anxiety to know more about it, Shyam explained about absorption of energy and re-emission of photons in the visible region. He also advised him not to touch or break any items in the lab for the thirst of knowledge. (a) What is the moral you derive from Shyam? (b) Which series in the hydrogen spectrum is in the visible region?
Answer: (a) Concern for his brother, and care for school property.
(b) The Balmer series.
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