Exercises — Atoms

1. 13 marksNCERT Exercises, Chapter 12

   Choose the correct alternative from the clues given at the end of each statement: (a) The size of the atom in Thomson's model is .......... the atomic size in Rutherford's model. (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (c) A classical atom based on .......... is doomed to collapse. (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (e) The positively charged part of the atom possesses most of the mass in ..........

   Hint. Contrast the two models on where the positive charge sits and whether the electron is at rest or in orbit.

   (a) no different from. Both models describe an atom of roughly $10^{-10}$ m. What differs is the arrangement of charge inside, not the overall size.

   (b) Thomson's model; Rutherford's model. In Thomson's model the electrons sit embedded in a uniform sphere of positive charge, where the net force on them is zero, so they are in stable equilibrium. In Rutherford's model the electron orbits the nucleus and is always being pulled towards it, so it always experiences a net force.

   (c) Rutherford's model. An orbiting electron is accelerating, and classical electromagnetism requires an accelerating charge to radiate energy continuously. Losing energy, it would spiral inward and reach the nucleus in about $10^{-8}$ s, so the classical Rutherford atom is unstable.

   (d) Thomson's model; Rutherford's model. Thomson spread the positive charge and mass smoothly through the whole atom. Rutherford concentrated almost all of it into a nucleus about $10^{-15}$ m across, leaving the rest of the atom essentially empty.

   (e) both the models. In each case the positive part carries nearly all the mass, since electrons are around 1836 times lighter than a proton. The models differ in how that positive matter is distributed, not in how much mass it carries.

   ✦ (a) no different from (b) Thomson's model; Rutherford's model (c) Rutherford's model (d) Thomson's model; Rutherford's model (e) both the models

2. 23 marksNCERT Exercises, Chapter 12

   Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

   Hint. Compare the mass and charge of a hydrogen nucleus with those of the alpha particle itself.

   Almost no large-angle scattering would be observed.

   The reason is mass. A hydrogen nucleus is a single proton of mass about 1 u, while an alpha particle has mass about 4 u. A light target cannot turn back a heavier projectile: in a collision the alpha particle would simply push the proton aside and carry on with very little deflection, much as a bowling ball is barely deflected by a marble.

   The charge matters too. Gold has $Z=79$ and hydrogen has $Z=1$, so the Coulomb repulsion driving the scattering is weaker by a factor of 79 at any given distance.

   Rutherford's key observation was that about one alpha particle in 8000 was scattered through more than $90^{\circ }$, and it was that rare backscattering which proved the existence of a small, massive, highly charged nucleus. With a hydrogen target the backscattering would essentially disappear, so the experiment would not have revealed the nuclear structure of the atom.

   ✦ Almost no large-angle scattering, since a hydrogen nucleus is much lighter than the alpha particle and carries only one unit of charge; the experiment would not have revealed the nucleus

3. 32 marksNCERT Exercises, Chapter 12

   A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

   Hint. Bohr's third postulate fixes the photon energy as the difference between the two levels.

   By Bohr's third postulate, the energy difference between the two levels is carried away entirely by a single photon:

   $h\nu =E_2-E_1$

   Converting the energy difference to joules:

   $\Delta E=2.3\times 1.6\times 10^{-19}=3.68\times 10^{-19}\text{J}$

   Therefore:

   $\nu =\frac{\Delta E}{h}=\frac{3.68\times 10^{-19}}{6.63\times 10^{-34}}=5.55\times 10^{14}\text{Hz}$

   This lies in the visible region, corresponding to a wavelength of about 540 nm, which is green light. The frequency is fixed entirely by the level spacing, since the photon must carry away exactly the energy the atom loses.

   ✦ nu = 5.55 x 10^14 Hz

4. 42 marksNCERT Exercises, Chapter 12

   The ground state energy of hydrogen atom is $-13.6$ eV. What are the kinetic and potential energies of the electron in this state?

   Hint. In any Bohr orbit the kinetic energy is minus the total energy, and the potential energy is twice the total energy.

   For an electron bound in a Coulomb field, the Bohr model gives fixed relationships between the three energies. Equating the Coulomb attraction to the centripetal requirement gives:

   $K=\frac{1}{2}\frac{e^2}{4\pi {\epsilon }_{0}r},U=-\frac{e^2}{4\pi {\epsilon }_{0}r}=-2K$

   So the total energy is $E=K+U=K-2K=-K$, which gives the two standard results:

   $K=-E,U=2E$

   Substituting $E=-13.6$ eV:

   $K=+13.6\text{eV},U=-27.2\text{eV}$

   A useful check: $K+U=13.6-27.2=-13.6$ eV, recovering the total energy. The negative total energy is what makes the atom bound, since 13.6 eV must be supplied to free the electron.

   ✦ Kinetic energy = +13.6 eV; potential energy = -27.2 eV

5. 53 marksNCERT Exercises, Chapter 12

   A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n=4$ level. Determine the wavelength and frequency of photon.

   Hint. Find the energy difference between the two levels using the Bohr energy formula, then convert it to a wavelength.

   The Bohr energy levels of hydrogen are:

   $E_n=-\frac{13.6}{n^2}\text{eV}$

   So $E_1=-13.6$ eV and $E_4=-\frac{13.6}{16}=-0.85$ eV. The energy absorbed is:

   $\Delta E=E_4-E_1=-0.85-(-13.6)=12.75\text{eV}$

   Converting to joules and finding the wavelength:

   $\lambda =\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times 3\times 10^8}{12.75\times 1.6\times 10^{-19}}=9.7\times 10^{-8}\text{m}$

   That is 97 nm, in the ultraviolet. The frequency follows:

   $\nu =\frac{c}{\lambda }=\frac{3\times 10^8}{9.7\times 10^{-8}}=3.09\times 10^{15}\text{Hz}$

   The shortcut $\lambda \text{(nm)}=1240/E\text{(eV)}$ gives the same answer in one step.

   ✦ lambda = 9.7 x 10^-8 m, about 97 nm; nu = 3.09 x 10^15 Hz

6. 64 marksNCERT Exercises, Chapter 12

   (a) Using the Bohr's model calculate the speed of the electron in a hydrogen atom in the $n=1$, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

   Hint. Speed falls as 1/n and radius grows as n squared, so work out how the period must scale.

   (a) Speeds. Bohr's quantisation condition $mvr=\frac{nh}{2\pi }$ combined with the force balance gives:

   $v_n=\frac{e^2}{2{\epsilon }_{0}nh}=\frac{2.18\times 10^6}{n}{\text{m s}}^{-1}$

   so the speed falls as $1/n$:

   $v_1=2.18\times 10^6,v_2=1.09\times 10^6,v_3=7.27\times 10^5{\text{m s}}^{-1}$

   Note that $v_1$ is about $c/137$, which is why the non-relativistic treatment is adequate.

   (b) Orbital periods. The radius grows as $r_n=0.53\times 10^{-10}{n}^2$ m, so:

   $T_n=\frac{2\pi r_n}{v_n}\propto \frac{n^2}{1/n}=n^3$

   The period therefore scales as $n^3$. Computing the first:

   $T_1=\frac{2\pi \times 0.53\times 10^{-10}}{2.18\times 10^6}=1.53\times 10^{-16}\text{s}$

   and scaling by $n^3$:

   $T_2=8T_1=1.22\times 10^{-15}\text{s},T_3=27T_1=4.13\times 10^{-15}\text{s}$

   An electron in the ground state completes roughly $6.6\times 10^{15}$ orbits per second.

   ✦ (a) 2.18 x 10^6, 1.09 x 10^6 and 7.27 x 10^5 m/s (b) 1.53 x 10^-16, 1.22 x 10^-15 and 4.13 x 10^-15 s, scaling as n cubed

7. 72 marksNCERT Exercises, Chapter 12

   The radius of the innermost electron orbit of a hydrogen atom is $5.3\times 10^{-11}$ m. What are the radii of the $n=2$ and $n=3$ orbits?

   Hint. The Bohr radius scales with the square of the principal quantum number.

   In the Bohr model the orbit radius grows as the square of the principal quantum number:

   $r_n=n^2{r}_1$

   where $r_1=5.3\times 10^{-11}$ m is the Bohr radius. Therefore:

   $r_2=2^2\times 5.3\times 10^{-11}=4\times 5.3\times 10^{-11}=2.12\times 10^{-10}\text{m}$

   $r_3=3^2\times 5.3\times 10^{-11}=9\times 5.3\times 10^{-11}=4.77\times 10^{-10}\text{m}$

   The orbits spread out rapidly, since the spacing between successive radii grows with $n$. This is why highly excited atoms are enormous compared with ground-state ones.

   ✦ r2 = 2.12 x 10^-10 m; r3 = 4.77 x 10^-10 m

8. 84 marksNCERT Exercises, Chapter 12

   A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

   Hint. Find the highest level the beam can reach, then list every downward transition possible from it.

   At room temperature the hydrogen atoms are in the ground state, $E_1=-13.6$ eV. After absorbing at most 12.5 eV the atom can reach at most:

   $-13.6+12.5=-1.1\text{eV}$

   Comparing with the Bohr levels $E_n=-13.6/n^2$: $E_3=-1.51$ eV and $E_4=-0.85$ eV. Reaching $n=4$ would need 12.75 eV, which exceeds the 12.5 eV available, while reaching $n=3$ needs only 12.09 eV. So the atom is excited at most to $n=3$.

   From $n=3$ three downward transitions are possible:

   Transition	$\Delta E$ (eV)	$\lambda$ (nm)	Series
   $3\to 1$	12.09	102.6	Lyman
   $2\to 1$	10.20	121.6	Lyman
   $3\to 2$	1.89	656.5	Balmer

   So the Lyman series appears in the ultraviolet and the Balmer series in the visible, the 656 nm line being the familiar red one.

   A note on the current edition. This question asks which series are emitted, but the chapter no longer names them: Lyman, Paschen, Brackett and Pfund return zero hits, and Balmer appears only once as an 1885 historical reference. The series names are supplied here because the exercise requires them.

   ✦ Excitation reaches n = 3, giving the Lyman series at 102.6 nm and 121.6 nm and the Balmer series at 656.5 nm

9. 93 marksNCERT Exercises, Chapter 12

   In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius $1.5\times 10^{11}$ m with orbital speed $3\times 10^4$ m/s. (Mass of earth $=6.0\times 10^{24}$ kg.)

   Hint. Apply Bohr's angular momentum quantisation to the Earth-Sun system and see how large n becomes.

   Bohr's second postulate quantises angular momentum:

   $mvr=\frac{nh}{2\pi }⟹n=\frac{2\pi mvr}{h}$

   Substituting the Earth's values:

   $n=\frac{2\pi \times 6.0\times 10^{24}\times 3\times 10^4\times 1.5\times 10^{11}}{6.63\times 10^{-34}}$

   The numerator is $2\pi \times 2.7\times 10^{40}=1.70\times 10^{41}$, so:

   $n=\frac{1.70\times 10^{41}}{6.63\times 10^{-34}}=2.6\times 10^{74}$

   What this enormous number means. Successive levels differ by one unit in $n$, so the fractional spacing between adjacent orbits is about $1/n\approx 10^{-74}$. The allowed orbits are therefore packed so closely that the motion is indistinguishable from a continuous one.

   This illustrates the correspondence principle: quantum results reproduce classical behaviour in the limit of large quantum numbers, which is why planetary orbits need no quantum treatment.

   ✦ n = 2.6 x 10^74, so large that the quantised orbits are indistinguishable from continuous classical motion

Solutions written by the tuition.in editorial team and checked against leph204.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.