PMC National MDCAT Chemistry Atomic Structure — Set 3

PMC National MDCAT Chemistry Atomic Structure — Set 3

1. Question 1

   Q1. The energy of an electron in the nth orbit of hydrogen is -13.6/n^2 eV. Calculate the energy of an electron in the 3rd orbit.

   • A) -1.51 eV
   • B) -3.4 eV
   • C) -6.04 eV
   • D) -12.09 eV

   Answer: -1.51 eV

   Explanation: Energy = -13.6/3^2 = -1.51 eV. Option B is for n=2.

2. Question 2

   Q2. Calculate the wavelength of a photon emitted when an electron jumps from n=4 to n=2 in a hydrogen atom. (R=1.097×10^7 m^-1)

   • A) 486 nm
   • B) 434 nm
   • C) 410 nm
   • D) 656 nm

   Answer: 486 nm

   Explanation: 1/λ = R(1/2^2 - 1/4^2), λ = 486 nm. Option B is for n=4 to n=2 in H-like atom with Z=2.

3. Question 3

   Q3. The ionization energy of He+ is 54.4 eV. What is the ionization energy of Li2+?

   • A) 122.4 eV
   • B) 108.8 eV
   • C) 217.6 eV
   • D) 163.2 eV

   Answer: 122.4 eV

   Explanation: IE ∝ Z^2, Li2+ has Z=3, so IE = 54.4 × (3/2)^2 = 122.4 eV. Option C is for a wrong assumption.

4. Question 4

   Q4. What is the radius of the 2nd orbit in a hydrogen atom? (a0 = 0.529 Å)

   • A) 1.058 Å
   • B) 2.116 Å
   • C) 0.529 Å
   • D) 4.232 Å

   Answer: 2.116 Å

   Explanation: r ∝ n^2, r2 = 0.529 × 2^2 = 2.116 Å. Option A is for n=1 of He+.

5. Question 5

   Q5. The frequency of a spectral line in the emission spectrum of hydrogen is 3.29 × 10^15 Hz. Identify the transition.

   • A) n=2 to n=1
   • B) n=3 to n=1
   • C) n=3 to n=2
   • D) n=4 to n=2

   Answer: n=2 to n=1

   Explanation: ΔE = hf, corresponds to Lyman-alpha line (n=2 to n=1). Option B is for a different frequency.

6. Question 6

   Q6. Calculate the de Broglie wavelength of an electron moving at 10% of the speed of light.

   • A) 0.024 nm
   • B) 0.242 nm
   • C) 0.048 nm
   • D) 0.012 nm

   Answer: 0.242 nm

   Explanation: λ = h/mv, v = 0.1c, λ = 0.242 nm. Option A is for v=c.

7. Question 7

   Q7. What is the maximum number of electrons in an atom with n=3, l=2?

   • A) 2
   • B) 5
   • C) 10
   • D) 18

   Answer: 10

   Explanation: l=2 means d-orbital, max 10 electrons. Option D is for n=3 total electrons.

8. Question 8

   Q8. The energy required to remove an electron from the ground state of He+ is 54.4 eV. Calculate the energy required to remove both electrons from He.

   • A) 79 eV
   • B) 108.8 eV
   • C) 54.4 eV + 24.6 eV
   • D) 108.8 eV + 24.6 eV

   Answer: 54.4 eV + 24.6 eV

   Explanation: First electron removal energy = 24.6 eV, second = 54.4 eV, total = 79 eV. Option D incorrectly adds IE of He+ twice.

9. Question 9

   Q9. For a hydrogen atom, the energy of the nth orbit is given by En = -13.6/n^2 eV. The energy difference between n=1 and n=2 is?

   • A) 10.2 eV
   • B) 12.09 eV
   • C) 3.4 eV
   • D) 1.89 eV

   Answer: 10.2 eV

   Explanation: ΔE = E2 - E1 = -3.4 - (-13.6) = 10.2 eV. Option B is E2.

10. Question 10

    Q10. The wavelength of the first line of the Balmer series is 656 nm. What is the wavelength of the second line?

    • A) 486 nm
    • B) 434 nm
    • C) 410 nm
    • D) 397 nm

    Answer: 486 nm

    Explanation: First line is n=3 to n=2, second line is n=4 to n=2, λ = 486 nm. Option B is for n=5 to n=2.

11. Question 11

    Q11. The velocity of an electron in the 1st orbit of hydrogen is 2.19 × 10^6 m/s. What is the velocity in the 3rd orbit?

    • A) 7.3 × 10^5 m/s
    • B) 2.19 × 10^6 m/s
    • C) 6.57 × 10^6 m/s
    • D) 1.095 × 10^7 m/s

    Answer: 7.3 × 10^5 m/s

    Explanation: v ∝ 1/n, v3 = v1/3 = 7.3 × 10^5 m/s. Option B is for n=1.

12. Question 12

    Q12. What is the value of the azimuthal quantum number for the 4d subshell?

    • A) 0
    • B) 1
    • C) 2
    • D) 3

    Answer: 2

    Explanation: d-orbital has l=2. Option A is for s-orbital.

13. Question 13

    Q13. Calculate the energy of a photon with a wavelength of 400 nm.

    • A) 3.1 eV
    • B) 2.48 eV
    • C) 4.1 eV
    • D) 1.55 eV

    Answer: 3.1 eV

    Explanation: E = hc/λ, E = 3.1 eV. Option B is for λ=500 nm.

14. Question 14

    Q14. The ionization energy of hydrogen is 13.6 eV. What is the ionization energy of Be3+?

    • A) 54.4 eV
    • B) 108.8 eV
    • C) 217.6 eV
    • D) 13.6 eV

    Answer: 217.6 eV

    Explanation: IE ∝ Z^2, Be3+ has Z=4, so IE = 13.6 × 4^2 = 217.6 eV. Option B is for Z=3 (Li2+).

15. Question 15

    Q15. What is the ratio of the radii of the 1st and 2nd orbits in a hydrogen atom?

    • A) 1:2
    • B) 1:4
    • C) 2:1
    • D) 4:1

    Answer: 1:4

    Explanation: r ∝ n^2, r1:r2 = 1^2:2^2 = 1:4. Option A is for a wrong proportionality.

16. Question 16

    Q16. The frequency of the radiation emitted when an electron falls from n=4 to n=1 in a hydrogen atom is?

    • A) 3.08 × 10^15 Hz
    • B) 2.47 × 10^15 Hz
    • C) 1.55 × 10^15 Hz
    • D) 3.29 × 10^15 Hz

    Answer: 3.08 × 10^15 Hz

    Explanation: ΔE = hf, corresponds to a specific transition frequency. Option D is for n=2 to n=1.

17. Question 17

    Q17. Calculate the number of nodes in the 3d orbital.

    • A) 0
    • B) 1
    • C) 2
    • D) 3

    Answer: 1

    Explanation: Number of nodes = n-l-1, for 3d, n=3, l=2, nodes = 0. Radial nodes = n-l-1 = 0, total nodes = 0 + angular nodes = 2, but radial nodes = 0. Option C is incorrect.

18. Question 18

    Q18. What is the maximum number of electrons that can be accommodated in the M shell?

    • A) 8
    • B) 18
    • C) 32
    • D) 50

    Answer: 18

    Explanation: M shell has n=3, max electrons = 2n^2 = 18. Option C is for N shell.

19. Question 19

    Q19. The wavelength of an electron moving with a velocity of 2.05 × 10^7 m/s is?

    • A) 0.355 Å
    • B) 3.55 Å
    • C) 35.5 Å
    • D) 355 Å

    Answer: 3.55 Å

    Explanation: λ = h/mv, λ = 3.55 Å (using h = 6.626 × 10^-34 and m = 9.11 × 10^-31).

20. Question 20

    Q20. The energy of the electron in the ground state of hydrogen is -13.6 eV. What is the energy in the ground state of Li2+?

    • A) -122.4 eV
    • B) -54.4 eV
    • C) -13.6 eV
    • D) -27.2 eV

    Answer: -122.4 eV

    Explanation: E ∝ Z^2, Li2+ has Z=3, E = -13.6 × 3^2 = -122.4 eV. Option B is for He+.