CHAPTER 12: ATOMS
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
- Physical Constants: $\displaystyle {R = 1.097 \times 10^7\text{ m}^{-1}}$, $\displaystyle {h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}}$, $\displaystyle {c = 3 \times 10^8\text{ m/s}}$, $\displaystyle {e = 1.6 \times 10^{-19}\text{ C}}$, $\displaystyle {m_e = 9.1 \times 10^{-31}\text{ kg}}$, $\displaystyle {a_0 = 53\text{ pm} = 0.53\text{ \AA}}$.
SECTION A (16 Marks)
Q.1. The magnetic field induction produced at the center of the orbit due to an electron revolving in the $n$-th Bohr orbit of a hydrogen atom is proportional to:
Explanation: Magnetic field at center $\displaystyle {B = {\frac{{\mu_0 I}}{{2 r}}} = {\frac{{\mu_0 e f}}{{2 r}}}}$. Since speed $\displaystyle {v \propto 1/n}$ and radius $\displaystyle {r \propto n^2}$, frequency $\displaystyle {f = v/(2\pi r) \propto 1/n^3}$. Thus $\displaystyle {B \propto (1/n^3) / n^2 = 1/n^5 = n^{-5}}$.
Q.2. When an electron in a hydrogen atom transitions from the fourth excited state ($n=5$) to the ground state ($n=1$):
Explanation: As $n$ decreases ($n=5 \to 1$), radius $r$ decreases. Kinetic energy $\displaystyle {K = +k e^2/(2r)}$ increases, while potential energy $\displaystyle {U = -k e^2/r}$ becomes more negative (decreases).
Q.3. The speed of an electron in the ground state orbit of a hydrogen atom is approximately:
Explanation: In ground state ($n=1$), electron orbital speed $\displaystyle {v_1 = {\frac{{e^2}}{{2 \varepsilon_0 h}}} \approx 2.18 \times 10^6\text{ m/s} \approx {\frac{{c}}{{137}}}}$.
Q.4. Taking the Bohr radius as $\displaystyle {a_0 = 53\text{ pm}}$, the radius of a $\displaystyle {\text{Li}^{2+}}$ ion ($Z=3$) in its ground state will be about:
Explanation: Radius $\displaystyle {r_n = a_0 {\frac{{n^2}}{{Z}}} = 53 \times {\frac{{1^2}}{{3}}} \approx 17.67\text{ pm} \approx 18\text{ pm}}$.
Q.5. Which of the following electronic transitions in a hydrogen atom emits photons of the lowest frequency?
Explanation: Photon frequency $\displaystyle {\nu \propto \Delta E = 13.6 \left( {\frac{{1}}{{n_f^2}}} - {\frac{{1}}{{n_i^2}}} \right)\text{ eV}}$. For $\displaystyle {n = 4 \to 3}$, $\displaystyle {\Delta E = 13.6 (1/9 - 1/16) = 0.66\text{ eV}}$, which is the smallest energy gap among options.
Q.6. Which of the following cannot be an ionization energy state value for a hydrogen-like system?
Explanation: Bound state energies are $\displaystyle {E_n = -13.6/n^2\text{ eV}}$. For $n=2, 3, 4$, $|E_n| = 3.4\text{ eV}, 1.51\text{ eV}, 0.85\text{ eV}$. $0.27\text{ eV}$ does not correspond to an integer $n$.
Q.7. How many revolutions per second does an electron complete in the first Bohr orbit of a hydrogen atom?
Explanation: Frequency $\displaystyle {f = {\frac{{v_1}}{{2\pi r_1}}} = {\frac{{2.18 \times 10^6}}{{2\pi \times (5.3 \times 10^{-11})}}} \approx 6.57 \times 10^{15}\text{ rev/s}}$.
Q.8. When an electron is excited to the $n$-th energy level in hydrogen, the maximum possible number of spectral lines emitted during de-excitation is:
Explanation: Total possible downward transitions from state $n$ to ground state is given by the combination formula $\displaystyle {^n C_2 = {\frac{{n(n-1)}}{{2}}}}$.
Q.9. The ratio of the energy of a hydrogen atom in its first excited state ($n=2$) to its second excited state ($n=3$) is:
Explanation: $\displaystyle {E_n \propto 1/n^2 \implies {\frac{{E_2}}{{E_3}}} = {\frac{{-13.6/4}}{{-13.6/9}}} = {\frac{{9}}{{4}}}}$.
Q.10. Bohr's atomic model is strictly applicable for:
Explanation: Bohr's model assumes a central Coulomb potential with a single orbiting electron ($\displaystyle {\text{H}, \text{He}^+, \text{Li}^{2+}}$) and cannot handle multi-electron inter-electronic repulsion.
Q.11. The total energy of an electron in the ground state of a hydrogen atom is $\displaystyle {-13.6\text{ eV}}$. Its kinetic energy in this state is:
Explanation: In a Coulomb field, Kinetic Energy $\displaystyle {K = -E_{\text{total}} = -(-13.6\text{ eV}) = +13.6\text{ eV}}$.
Q.12. The speed of an electron in the 4th orbit ($n=4$) of a hydrogen atom is:
Explanation: $\displaystyle {v_n = v_1 / n = (c/137) / 4 = c / 548}$.
Directions for Q.13 to Q.16:
(a) Both Assertion and Reason are true and Reason is correct explanation of A.
(b) Both Assertion and Reason are true but Reason is NOT correct explanation of A.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
Q.13. Assertion (A): According to Bohr's atomic model, the ratio of orbital angular momenta of an electron in the first excited state to the ground state is $\displaystyle {2 : 1}$.
Reason (R): In a Bohr atom, the angular momentum of an orbiting electron is directly proportional to the principal quantum number $\displaystyle {n}$.
Explanation: $\displaystyle {L = n h / (2\pi) \implies L \propto n}$. For first excited state ($n=2$) vs ground state ($n=1$), $\displaystyle {L_2/L_1 = 2/1}$.
Q.14. Assertion (A): The positively charged nucleus of an atom has a radius of the order of $\displaystyle {10^{-15}\text{ m}}$.
Reason (R): In the $\alpha$-particle scattering experiment, the distance of closest approach for energetic $\alpha$-particles is of the order of $\displaystyle {10^{-14}\text{ m} - 10^{-15}\text{ m}}$.
Q.15. Assertion (A): According to classical electromagnetic theory, the trajectory of an electron spiraling into the nucleus in Rutherford's model is continuous.
Reason (R): An accelerating electric charge continuously radiates electromagnetic energy at the frequency of its revolution.
Q.16. Assertion (A): The simple Bohr model of the hydrogen atom fails to explain the fine structure of spectral lines.
Reason (R): The simple Bohr model treats orbits non-relativistically and neglects electron spin and spin-orbit interaction.
SECTION B (10 Marks)
Q.17. The radius of the innermost electron orbit of a hydrogen atom is $\displaystyle {5.3 \times 10^{-11}\text{ m}}$. What is the radius of the orbit in the second excited state?
$\displaystyle {r_3 = n^2 r_1 = (3)^2 \times (5.3 \times 10^{-11}\text{ m}) = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10}\text{ m} = 0.477\text{ nm}}$.
Q.18. Which spectral series of the hydrogen atom lies completely in the ultraviolet region, and why?
An electron in a hydrogen atom is in its second excited state. Calculate the wavelength of the spectral lines in the Lyman series emitted during permissible transitions ($\displaystyle {R = 1.1 \times 10^7\text{ m}^{-1}}$).
The Lyman series ($\displaystyle {n_f = 1}$) lies completely in the UV region because transitions to the ground state involve the largest energy gaps ($\displaystyle {\Delta E \ge 10.2\text{ eV}}$), producing high-frequency ultraviolet photons.
OR Part: Second excited state means $\displaystyle {n = 3}$. Permissible Lyman transitions: $\displaystyle {3 \to 1}$ and $\displaystyle {2 \to 1}$.
(1) $\displaystyle {3 \to 1}$: $\displaystyle {1/\lambda = R(1/1^2 - 1/3^2) = R(8/9) \implies \lambda = {\frac{{9}}{{8 R}}} = {\frac{{9}}{{8 \times 1.1 \times 10^7}}} \approx 102.3\text{ nm}}$.
(2) $\displaystyle {2 \to 1}$: $\displaystyle {1/\lambda = R(1/1^2 - 1/2^2) = R(3/4) \implies \lambda = {\frac{{4}}{{3 R}}} = {\frac{{4}}{{3 \times 1.1 \times 10^7}}} \approx 121.5\text{ nm}}$.
Q.19. An energy level diagram of an element shows energy states $\displaystyle {-13.6\text{ eV}, -3.4\text{ eV}, -1.5\text{ eV}, -0.85\text{ eV}}$. Identify which transition corresponds to emission of a photon of wavelength $\displaystyle {102.7\text{ nm}}$.
Energy difference between levels $\displaystyle {n=3}$ ($\displaystyle {-1.5\text{ eV}}$) and $\displaystyle {n=1}$ ($\displaystyle {-13.6\text{ eV}}$) is:
$\displaystyle {\Delta E = -1.5\text{ eV} - (-13.6\text{ eV}) = 12.1\text{ eV}}$.
Thus, Transition D ($\displaystyle {n=3 \to n=1}$) corresponds to $\displaystyle {102.7\text{ nm}}$.
Q.20. The short wavelength limit for the Lyman series of hydrogen is $\displaystyle {913.4\text{ \AA}}$. Calculate the short wavelength limit for the Balmer series.
Lyman series limit ($\displaystyle {n_f = 1}$): $\displaystyle {1/\lambda_L = R(1/1^2 - 0) = R \implies \lambda_L = 1/R = 913.4\text{ \AA}}$.
Balmer series limit ($\displaystyle {n_f = 2}$): $\displaystyle {1/\lambda_B = R(1/2^2 - 0) = R/4 \implies \lambda_B = 4/R = 4 \lambda_L}$.
$\displaystyle {\lambda_B = 4 \times 913.4\text{ \AA} = 3653.6\text{ \AA} \approx 365.4\text{ nm}}$.
Q.21. Find the ratio between the wavelengths of the most energetic spectral lines in the Balmer and Paschen series of the hydrogen spectrum.
Balmer series ($\displaystyle {n_f = 2}$): $\displaystyle {1/\lambda_B = R(1/2^2 - 0) = R/4 \implies \lambda_B = 4/R}$.
Paschen series ($\displaystyle {n_f = 3}$): $\displaystyle {1/\lambda_P = R(1/3^2 - 0) = R/9 \implies \lambda_P = 9/R}$.
Ratio $\displaystyle {\lambda_P / \lambda_B = (9/R) / (4/R) = 9 : 4}$.
SECTION C (21 Marks)
Q.22. (a) Calculate the radius of the $n=2$ orbit of a hydrogen atom given $r_1 = 5.3 \times 10^{-11}\text{ m}$.
(b) The total energy of an electron in the second excited state of a hydrogen atom is $\displaystyle {-1.51\text{ eV}}$. Find its kinetic energy and potential energy in this state.
A $12.5\text{ eV}$ electron beam bombards gaseous hydrogen at room temperature. Up to which energy level are hydrogen atoms excited? Calculate the wavelength of the first member of the Lyman series.
(a) $\displaystyle {r_2 = 2^2 \times r_1 = 4 \times (5.3 \times 10^{-11}\text{ m}) = 21.2 \times 10^{-11}\text{ m} = 0.212\text{ nm}}$.
(b) For total energy $\displaystyle {E = -1.51\text{ eV}}$:
(i) Kinetic Energy $\displaystyle {K = -E = +1.51\text{ eV}}$.
(ii) Potential Energy $\displaystyle {U = 2 E = 2 \times (-1.51\text{ eV}) = -3.02\text{ eV}}$.
OR Part: Ground state $\displaystyle {E_1 = -13.6\text{ eV}}$. Energy after absorbing $\displaystyle {12.5\text{ eV}}$: $\displaystyle {E' = -13.6 + 12.5 = -1.1\text{ eV}}$.
Energy levels: $\displaystyle {E_1 = -13.6\text{ eV}, E_2 = -3.4\text{ eV}, E_3 = -1.51\text{ eV}, E_4 = -0.85\text{ eV}}$.
Since $\displaystyle {-1.1\text{ eV} > -1.51\text{ eV}}$ but $\displaystyle {< -0.85\text{ eV}}$, atom is excited up to $\displaystyle {n = 3}$ level (second excited state).
First member of Lyman ($\displaystyle {2 \to 1}$): $\displaystyle {\lambda = {\frac{4}{3 R}} = {\frac{4}{3 \times 1.097 \times 10^7}} \approx 121.5\text{ nm}}$.
Q.23. Using Bohr's postulates, obtain the expression for the total energy of an electron in stationary orbits of a hydrogen atom. Draw an energy level diagram showing the Balmer series transitions.
Kinetic Energy $\displaystyle {K = {\frac{1}{2}} m v^2 = {\frac{{e^2}}{{8\pi \varepsilon_0 r}}}}$. Potential Energy $\displaystyle {U = -{\frac{{e^2}}{{4\pi \varepsilon_0 r}}}}$.
Total Energy $\displaystyle {E = K + U = -{\frac{{e^2}}{{8\pi \varepsilon_0 r}}}}$.
Substituting Bohr quantized radius $\displaystyle {r_n = {\frac{{\varepsilon_0 n^2 h^2}}{{\pi m e^2}}}}$ yields: $\displaystyle {E_n = -{\frac{{m e^4}}{{8 \varepsilon_0^2 n^2 h^2}}} = -{\frac{{13.6}}{{n^2}}}\text{ eV}}$.
Q.24. An electron transitions from energy level $\displaystyle {-0.85\text{ eV}}$ to $\displaystyle {-3.4\text{ eV}}$ in a hydrogen atom. Calculate the wavelength of the emitted photon and name the spectral series.
Emitted photon energy: $\displaystyle {\Delta E = E_i - E_f = -0.85 - (-3.4) = 2.55\text{ eV} = 4.08 \times 10^{-19}\text{ J}}$.
Wavelength: $\displaystyle {\lambda = {\frac{{hc}}{{\Delta E}}} = {\frac{{6.63 \times 10^{-34} \times 3 \times 10^8}}{{4.08 \times 10^{-19}}}} = 4.875 \times 10^{-7}\text{ m} = 487.5\text{ nm} = 4875\text{ \AA}}$.
Since final level is $\displaystyle {n_f = 2}$, this belongs to the Balmer series (visible region).
Q.25. In a Geiger–Marsden experiment, calculate the distance of closest approach for an $\alpha$-particle of $8\text{ MeV}$ impinging on a gold nucleus ($Z=80$). How is this distance affected if kinetic energy is doubled?
$\displaystyle {K_\alpha = {\frac{{1}}{{4\pi \varepsilon_0}}} {\frac{{(2e)(Ze)}}{{r_0}}} \implies r_0 = {\frac{{1}}{{4\pi \varepsilon_0}}} {\frac{{2 Z e^2}}{{K_\alpha}}}}$.
For $\displaystyle {Z=80, K_\alpha = 8\text{ MeV} = 8 \times 10^6 \times 1.6 \times 10^{-19}\text{ J} = 1.28 \times 10^{-12}\text{ J}}$:
$\displaystyle {r_0 = {\frac{{(9 \times 10^9) \times 2 \times 80 \times (1.6 \times 10^{-19})^2}}{{1.28 \times 10^{-12}}}} = {\frac{{3.6864 \times 10^{-26}}}{{1.28 \times 10^{-12}}}} = 2.88 \times 10^{-14}\text{ m} = 28.8\text{ fm}}$.
Since $\displaystyle {r_0 \propto 1/K_\alpha}$, doubling kinetic energy halves the distance of closest approach ($\displaystyle {r'_0 = r_0/2 = 14.4\text{ fm}}$).
Q.26. (a) Using Bohr's second postulate, show that the circumference of the $n$-th orbit is $n$ times the de Broglie wavelength of the electron.
(b) If an electron is in the third excited state ($n=4$), find the maximum number of spectral lines emitted upon de-excitation to ground state.
(a) Bohr's angular momentum quantization: $\displaystyle {m v r = {\frac{{n h}}{{2\pi}}}} \implies 2\pi r = n {\left({\frac{{h}}{{m v}}}\right)}$.
By de Broglie relation, $\displaystyle {\lambda = {\frac{{h}}{{m v}}}}$. Substituting gives $\displaystyle {2\pi r = n \lambda}$. Thus, orbit circumference equals $n$ de Broglie wavelengths.
(b) Third excited state corresponds to $\displaystyle {n = 4}$. Maximum spectral lines $\displaystyle {N = {\frac{{n(n-1)}}{{2}}} = {\frac{{4 \times 3}}{{2}}} = 6}$ lines.
Q.27. Ground state energy of hydrogen is $\displaystyle {-13.6\text{ eV}}$.
(i) What is the potential energy of the electron in the 3rd excited state?
(ii) If the electron jumps from 3rd excited state to ground state, calculate emitted photon wavelength.
Total energy $\displaystyle {E_4 = -13.6/4^2 = -13.6/16 = -0.85\text{ eV}}$.
(i) Potential Energy $\displaystyle {U_4 = 2 E_4 = 2 \times (-0.85\text{ eV}) = -1.70\text{ eV}}$.
(ii) Transition $\displaystyle {4 \to 1}$: $\displaystyle {\Delta E = -0.85 - (-13.6) = 12.75\text{ eV}}$.
Wavelength $\displaystyle {\lambda = {\frac{{1240\text{ eV}\cdot\text{nm}}}{{12.75\text{ eV}}}} \approx 97.25\text{ nm} = 972.5\text{ \AA}}$.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.28. Case Study 1: Geiger–Marsden $\alpha$-Scattering Experiment & Nuclear Dimensions.
In 1911, Geiger and Marsden directed $\alpha$-particles ($\displaystyle {q = +2e}$) at a thin gold foil ($\displaystyle {Z=79}$). Most $\alpha$-particles passed undeflected, but about 1 in 8000 was deflected by $\displaystyle {> 90^\circ}$.
(i) Fact that most $\alpha$-particles pass undeflected implies:
(a) Atom is dense solid (b) Atom is mostly empty space (c) Nucleus is neutral
(ii) Large-angle deflection occurs due to:
(a) Collision with electrons (b) Intense Coulomb repulsion from concentrated positive nucleus
(iii) Impact parameter $b$ for an $\alpha$-particle scattered through $\displaystyle {\theta = 180^\circ}$ is:
(a) Maximum (b) Zero (c) Infinity
(iv) Ratio of atomic radius ($\displaystyle {\sim 10^{-10}\text{ m}}$) to nuclear radius ($\displaystyle {\sim 10^{-15}\text{ m}}$) is:
(a) $\displaystyle {10^2}$ (b) $\displaystyle {10^5}$ (c) $\displaystyle {10^{10}}$
Q.29. Case Study 2: Atomic Hydrogen Spectral Series.
When an electron jumps from higher level $n_i$ to lower level $n_f$, a photon is emitted with wave number $\displaystyle {1/\lambda = R(1/n_f^2 - 1/n_i^2)}$.
(i) Spectral series obtained for transitions ending at $n_f=3$ is:
(a) Lyman (b) Balmer (c) Paschen (d) Brackett
(ii) Balmer series lines lie in which region of the EM spectrum?
(a) Ultraviolet (b) Visible (c) Infrared (d) X-ray
(iii) Maximum energy of a photon emitted in the hydrogen emission spectrum is:
(a) $\displaystyle {13.6\text{ eV}}$ (b) $\displaystyle {10.2\text{ eV}}$ (c) $\displaystyle {3.4\text{ eV}}$
(iv) Minimum energy required to ionize a hydrogen atom from its first excited state ($n=2$) is:
(a) $\displaystyle {13.6\text{ eV}}$ (b) $\displaystyle {10.2\text{ eV}}$ (c) $\displaystyle {3.4\text{ eV}}$ (d) $\displaystyle {1.51\text{ eV}}$
SECTION E (15 Marks)
Q.30. (a) Draw a schematic arrangement of the Geiger–Marsden experiment. Explain trajectory of $\alpha$-particles in the Coulomb field of a target nucleus and define impact parameter.
(b) Estimate distance of closest approach for a $\displaystyle {7.7\text{ MeV}}$ $\alpha$-particle impinging on a gold nucleus ($Z=80$).
(i) State Bohr's quantization condition for stationary orbits and explain it using de Broglie's hypothesis.
(ii) Three energy levels A, B, C satisfy $\displaystyle {E_C > E_B > E_A}$. Radiations emitted in transitions $C \to B, B \to A, C \to A$ have wavelengths $\displaystyle {\lambda_1, \lambda_2, \lambda_3}$. Find the relation between $\displaystyle {\lambda_1, \lambda_2, \lambda_3}$.
(a) Impact parameter $b$ is the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the center of the nucleus. Large $b \implies$ small deflection; $b \approx 0 \implies$ head-on collision ($\displaystyle {\theta \approx 180^\circ}$).
(b) $\displaystyle {K_\alpha = 7.7\text{ MeV} = 7.7 \times 1.6 \times 10^{-13}\text{ J} = 1.232 \times 10^{-12}\text{ J}}$.
$\displaystyle {r_0 = {\frac{{(9 \times 10^9) \times 2 \times 80 \times (1.6 \times 10^{-19})^2}}{{1.232 \times 10^{-12}}}} \approx 3.0 \times 10^{-14}\text{ m} = 30\text{ fm}}$.
OR Part (ii): Energy relation: $\displaystyle {E_{C \to A} = E_{C \to B} + E_{B \to A}}$.
$\displaystyle {{\frac{{hc}}{{\lambda_3}}} = {\frac{{hc}}{{\lambda_1}}} + {\frac{{hc}}{{\lambda_2}}} \implies {\frac{{1}}{{\lambda_3}}} = {\frac{{1}}{{\lambda_1}}} + {\frac{{1}}{{\lambda_2}}} \implies \lambda_3 = {\frac{{\lambda_1 \lambda_2}}{{\lambda_1 + \lambda_2}}}}$.
Q.31. (i) A hydrogen atom initially in the ground state absorbs a photon which excites it to the $n=4$ level. Determine the wavelength of the absorbed photon.
(ii) Find the radius of the $n=4$ orbit given ground state radius $r_1 = 5.3 \times 10^{-11}\text{ m}$.
Find the transition in hydrogen emission spectrum that yields a photon of wavelength $\displaystyle {496\text{ nm}}$ and specify the transition yielding maximum wavelength.
(i) Energy required: $\displaystyle {\Delta E = E_4 - E_1 = -0.85\text{ eV} - (-13.6\text{ eV}) = 12.75\text{ eV}}$.
Wavelength: $\displaystyle {\lambda = {\frac{{hc}}{{\Delta E}}} = {\frac{{1240\text{ eV}\cdot\text{nm}}}{{12.75\text{ eV}}}} \approx 97.25\text{ nm} = 970\text{ \AA}}$.
(ii) Radius: $\displaystyle {r_4 = 4^2 \times r_1 = 16 \times 5.3 \times 10^{-11}\text{ m} = 8.48 \times 10^{-10}\text{ m} = 8.48\text{ \AA}}$.
OR Part: Wavelength $\displaystyle {496\text{ nm}}$ falls in visible spectrum (Balmer, $n_f=2$).
$\displaystyle {1/496 = R(1/4 - 1/n_i^2) \implies n_i = 4}$. Thus, transition is $\displaystyle {n=4 \to n=2}$.
Maximum wavelength corresponds to minimum energy gap in Balmer: $\displaystyle {n=3 \to n=2}$ ($\displaystyle {\lambda = 656\text{ nm}}$).
Q.32. (a) Define ionization energy. How would the ionization energy change if the electron in a hydrogen atom were replaced by a muon ($\mu^-$) of mass $200 m_e$?
(b) Derive expression for distance of closest approach $r_0$ for an $\alpha$-particle of kinetic energy $K$.
(a) Ionization Energy: Minimum energy required to completely remove an electron from its ground state to infinity ($\displaystyle {n=\infty}$).
From Bohr formula, binding energy $\displaystyle {E_1 = {\frac{{m e^4}}{{8 \varepsilon_0^2 h^2}}} \propto m}$. Replacing electron with a muon of mass $\displaystyle {m_\mu = 200 m_e}$ increases the ionization energy by 200 times ($\displaystyle {E_1' = 200 \times 13.6\text{ eV} = 2720\text{ eV} = 2.72\text{ keV}}$).
(b) Conservation of energy at turning point $\displaystyle {r_0}$: Initial $\displaystyle {K_\alpha = \text{Electrostatic PE} = {\frac{{1}}{{4\pi \varepsilon_0}}} {\frac{{(2e)(Ze)}}{{r_0}}} \implies r_0 = {\frac{{2 Z e^2}}{{4\pi \varepsilon_0 K_\alpha}}}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. In an experiment on $\alpha$-particle scattering by gold foil, closest approach is $30\text{ fm}$. If $\alpha$-particle speed is doubled, closest approach becomes:
Explanation: $\displaystyle {r_0 \propto 1/K \propto 1/v^2}$. Doubling $v$ increases $K$ fourfold, reducing $r_0$ to $\displaystyle {30/4 = 7.5\text{ fm}}$.
2. The ratio of orbital angular momenta of an electron in 1st excited state ($n=2$) to 2nd excited state ($n=3$) is:
3. The radius of the second Bohr orbit of singly ionized helium ($\text{He}^+, Z=2$) is:
Explanation: $\displaystyle {r = 0.529 \times (n^2/Z) = 0.529 \times (4/2) = 1.058\text{ \AA} \approx 1.06\text{ \AA}}$.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): The total energy of an electron revolving in any stationary orbit of a hydrogen atom is negative.
Reason (R): The electron is bound to the nucleus by the attractive Coulomb force.
2. Assertion (A): A hydrogen atom has only one electron, yet its emission spectrum contains numerous lines.
Reason (R): A gas sample contains a vast number of hydrogen atoms whose electrons sit in various excited states and undergo different transitions.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Bohr Model Parameters for Hydrogen-Like Ions
For hydrogen-like ions of atomic number $Z$, orbit radius is $\displaystyle {r_n = 0.53 n^2/Z\text{ \AA}}$ and energy is $\displaystyle {E_n = -13.6 Z^2/n^2\text{ eV}}$.
1. Ground state energy of $\text{He}^+$ ($Z=2$): (a) $\displaystyle {-13.6\text{ eV}}$ (b) $\displaystyle {-54.4\text{ eV}}$
2. Ratio of radius of $n=3$ orbit of $\text{He}^+$ to $n=2$ orbit of H: (a) $\displaystyle {9 : 8}$ (b) $\displaystyle {8 : 9}$
3. Energy needed to ionize $\text{Li}^{2+}$ ($Z=3$) from ground state: (a) $\displaystyle {122.4\text{ eV}}$ (b) $\displaystyle {13.6\text{ eV}}$
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
Radius: $\displaystyle {r_n = a_0 {\frac{{n^2}}{{Z}}} = 0.529 {\frac{{n^2}}{{Z}}}\text{ \AA}}$ | Velocity: $\displaystyle {v_n = v_1 {\frac{{Z}}{{n}}} = 2.18 \times 10^6 {\frac{{Z}}{{n}}}\text{ m/s}}$
Energy: $\displaystyle {E_n = -13.6 {\frac{{Z^2}}{{n^2}}}\text{ eV}}$ | $\displaystyle {K_n = -E_n = +13.6 {\frac{{Z^2}}{{n^2}}}\text{ eV}}$ | $\displaystyle {U_n = 2 E_n = -27.2 {\frac{{Z^2}}{{n^2}}}\text{ eV}}$
CONCEPTUAL SHORT QUESTIONS
1. Why does the classical Rutherford atomic model lead to the collapse of the atom?
2. Why are absorption spectrum lines of hydrogen gas at room temperature composed almost entirely of the Lyman series?
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