Unit 2: Atomic Structure — Online MCQ Test
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This chapter covers: Nature of electromagnetic radiation photoelectric effect spectrum of the hydrogen atom Bohr model of ahydrogen atom - its postulates derivation of the relations for the energy of t....
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Unit 2: Atomic Structure — Important Questions & Answers
Radius of the first excited state of Helium ion is given as x times radius of first stationary state of hydrogen atom. The value of x is:
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2
Radius formula r = a0 * n^2 / Z. For H ground state (n=1, Z=1), r1 = a0. For He+ first excited state (n=2, Z=2), r2 = a0 * (2^2) / 2 = 2*a0. Thus x = 2.
Radius formula r = a0 * n^2 / Z. For H ground state (n=1, Z=1), r1 = a0. For He+ first excited state (n=2, Z=2), r2 = a0 * (2^2) / 2 = 2*a0. Thus x = 2.
Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series The correct arrangement of the above lines in ascending order of energy is:
- A. (1) D < C < A < B
- B. (2) A < B < C < D
- C. (3) C < D < B < A
- D. (4) D < A < C < B
Answer: D. (4) D < A < C < B
The energy of a transition is proportional to (1/n1^2 - 1/n2^2). (A) n1=3, n2=4: ΔE ∝ (1/9 - 1/16) = 7/144 ≈ 0.0486. (B) n1=2, n2=4: ΔE ∝ (1/4 - 1/16) = 3/16 = 0.1875. (C) n1=3, n2=6: ΔE ∝ (1/9 - 1/36) = 3/36 = 0.0833. (D) n1=4, n2=8: ΔE ∝ (1/16 - 1/64) = 3/64 = 0.0468. Order: D(0.0468) < A(0.0486) < C(0.0833) < B(0.1875).
The energy of a transition is proportional to (1/n1^2 - 1/n2^2). (A) n1=3, n2=4: ΔE ∝ (1/9 - 1/16) = 7/144 ≈ 0.0486. (B) n1=2, n2=4: ΔE ∝ (1/4 - 1/16) = 3/16 = 0.1875. (C) n1=3, n2=6: ΔE ∝ (1/9 - 1/36) = 3/36 = 0.0833. (D) n1=4, n2=8: ΔE ∝ (1/16 - 1/64) = 3/64 = 0.0468. Order: D(0.0468) < A(0.0486) < C(0.0833) < B(0.1875).
The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is:
- A. x^2
- B. x/1.35
- C. 2x
- D. 1.35x
Answer: D. 1.35x
First Balmer line: n=3 to n=2 (dE1 proportional to 1/4 - 1/9 = 5/36). Second Balmer line: n=4 to n=2 (dE2 proportional to 1/4 - 1/16 = 3/16). Ratio dE2/dE1 = (3/16) / (5/36) = (3*36)/(16*5) = 108/80 = 1.35. Thus, E2 = 1.35x.
First Balmer line: n=3 to n=2 (dE1 proportional to 1/4 - 1/9 = 5/36). Second Balmer line: n=4 to n=2 (dE2 proportional to 1/4 - 1/16 = 3/16). Ratio dE2/dE1 = (3/16) / (5/36) = (3*36)/(16*5) = 108/80 = 1.35. Thus, E2 = 1.35x.
The wavelength of spectral line obtained in the spectrum of Li2+ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is
- A. 2.28 * 10^-7 cm
- B. 2.28 * 10^-6 cm
- C. 1.14 * 10^-7 cm
- D. 1.14 * 10^-6 cm
Answer: D. 1.14 * 10^-6 cm
n1 + n2 = 4 and n2 - n1 = 2 gives n2=3, n1=1. Using Rydberg formula: 1/lambda = R * Z^2 * (1/n1^2 - 1/n2^2). With Z=3, 1/lambda = R * 9 * (1/1 - 1/9) = R * 9 * 8/9 = 8R. Lambda = 1/(8R) = 1/(8 * 1.1*10^5) ≈ 1.14 * 10^-6 cm.
n1 + n2 = 4 and n2 - n1 = 2 gives n2=3, n1=1. Using Rydberg formula: 1/lambda = R * Z^2 * (1/n1^2 - 1/n2^2). With Z=3, 1/lambda = R * 9 * (1/1 - 1/9) = R * 9 * 8/9 = 8R. Lambda = 1/(8R) = 1/(8 * 1.1*10^5) ≈ 1.14 * 10^-6 cm.
Number of water molecules in washing soda and soda ash respectively are:
- A. 10 and 1
- B. 1 and 10
- C. 10 and 0
- D. 1 and 0
Answer: C. 10 and 0
Washing soda is sodium carbonate decahydrate, $\mathrm{Na_2CO_3\cdot 10H_2O}$, so it contains 10 water molecules of crystallization. Soda ash is anhydrous sodium carbonate, $\mathrm{Na_2CO_3}$, so it contains 0 water molecules. Hence the answer is 10 and 0.
Washing soda is sodium carbonate decahydrate, $\mathrm{Na_2CO_3\cdot 10H_2O}$, so it contains 10 water molecules of crystallization. Soda ash is anhydrous sodium carbonate, $\mathrm{Na_2CO_3}$, so it contains 0 water molecules. Hence the answer is 10 and 0.