Structure of Atom

What you'll learn

A clear, exam-ready guide to atomic structure for JEE — subatomic particles, Thomson to Bohr models, the hydrogen spectrum formulas, quantum numbers, and the Aufbau, Pauli and Hund rules.

This chapter has

Structure of Atom — solved practice questions

8 JEE Chemistry questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   The maximum number of electrons that can be accommodated in the shell with principal quantum number n = 3 is:

   • A8
   • B9
   • C18
   • D32
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   Correct answer: (C) 18

   Maximum electrons in a shell = 2n^2 = 2 x 3^2 = 18.

2. Q2easy

   The total number of orbitals present in the n = 3 shell is:

   • A3
   • B9
   • C5
   • D18
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   Correct answer: (B) 9

   Number of orbitals in a shell = n^2 = 3^2 = 9 (one 3s, three 3p, five 3d).

3. Q3easy

   According to Bohr's model, the radius of the third orbit (n = 3) of the hydrogen atom is: (radius of first orbit = 0.529 Angstrom)

   • A0.529 Angstrom
   • B1.587 Angstrom
   • C2.645 Angstrom
   • D4.76 Angstrom
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   Correct answer: (D) 4.76 Angstrom

   In Bohr's model r_n = 0.529 x n^2 / Z Angstrom. For H (Z=1), r_3 = 0.529 x 9 = 4.76 Angstrom.

4. Q4medium

   The energy of an electron in the second orbit (n = 2) of the hydrogen atom is: (Energy of ground state = -13.6 eV)

   • A-3.4 eV
   • B-6.8 eV
   • C-1.51 eV
   • D-13.6 eV
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   Correct answer: (A) -3.4 eV

   E_n = -13.6/n^2 eV. For n = 2, E = -13.6/4 = -3.4 eV.

5. Q5medium

   The ionization energy of the He+ ion (Z = 2) in its ground state is: (Ionization energy of hydrogen = 13.6 eV)

   • A13.6 eV
   • B27.2 eV
   • C54.4 eV
   • D6.8 eV
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   Correct answer: (C) 54.4 eV

   For a hydrogen-like species, ionization energy = 13.6 x Z^2 eV. For He+, IE = 13.6 x 4 = 54.4 eV.

6. Q6medium

   Which set of quantum numbers is NOT permissible for an electron in an atom?

   • An = 3, l = 2, m = -2, s = +1/2
   • Bn = 2, l = 2, m = 0, s = -1/2
   • Cn = 4, l = 0, m = 0, s = +1/2
   • Dn = 3, l = 1, m = +1, s = -1/2
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   Correct answer: (B) n = 2, l = 2, m = 0, s = -1/2

   For a given n, l ranges from 0 to n-1. The set n = 2, l = 2 is not allowed because l must be less than n (l can only be 0 or 1 when n = 2).

7. Q7hard

   The wavelength of the second line of the Balmer series (transition n = 4 to n = 2) of the hydrogen spectrum is approximately: (Rydberg constant R = 1.097 x 10^7 per metre)

   • A656 nm
   • B410 nm
   • C434 nm
   • D486 nm
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   Correct answer: (D) 486 nm

   1/lambda = R(1/2^2 - 1/4^2) = 1.097 x 10^7 x (1/4 - 1/16) = 1.097 x 10^7 x 3/16, giving lambda = 486 nm.

8. Q8hard

   The de Broglie wavelength associated with an electron accelerated through a potential difference of 100 V is approximately:

   • A0.123 nm
   • B1.23 nm
   • C0.012 nm
   • D12.3 nm
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   Correct answer: (A) 0.123 nm

   lambda = h / sqrt(2meV) = 12.27/sqrt(V) Angstrom = 12.27/10 = 1.227 Angstrom, i.e. about 0.123 nm.

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