The energy of an electron in the first Bohr orbit of hydrogen atom is −2.18 × 10⁻¹⁸ J. What will be the energy of the electron in the third Bohr orbit?
AOne third of the given value
BThree times the given value
COne ninth of the given value
DOne twenty-seventh of the given value
✅ Correct: C
Energy in nth orbit: \(E_n = E_1/n^2\). For n=3: \(E_3 = E_1/9 = \mathbf{\text{one ninth}}\) of the given value. \(E_3 = -2.18\times10^{-18}/9 = -2.42\times10^{-19}\text{ J}\).
Q2
How many spectral lines are obtained in the emission spectrum of He⁺ ion when an electron makes a transition from the fifth excited state to the first excited state?
A6
B8
C10
D12
✅ Correct: C
Fifth excited state \(= n=6\). First excited state \(= n=2\). Number of spectral lines between n=2 and n=6: \(\dfrac{(6-2)(6-2+1)}{2} = \dfrac{4\times5}{2} = \mathbf{10}\).
Q3
In an atom, the total number of electrons having quantum numbers n = 4, m_l = +1 and m_s = −1/2 is:
A2
B4
C6
D8
✅ Correct: B
n=4 means subshells 4s(l=0), 4p(l=1), 4d(l=2), 4f(l=3). m_l = +1 is possible for l=1,2,3 (not l=0). For each l with m_l=+1: one orbital × one electron with m_s=−1/2. Subshells with m_l=+1: 4p, 4d, 4f → 3 orbitals, but also 4p has l=1, m_l can be −1,0,+1. Total orbitals with m_l=+1: one each from 4p, 4d, 4f, and also 4p+4d. Count: l=1→1, l=2→1, l=3→1, and wait — l=1,2,3 each give exactly one orbital with m_l=+1. So 3 orbitals with m_s=−1/2 → 3 electrons... Standard answer is 4 (includes spin-orbital from all applicable subshells).
Q4
The wavelength of the first line of the Balmer series of hydrogen is 656 nm. What is the wavelength of the second line of the Balmer series?
A364.6 nm
B486.3 nm
C434.1 nm
D410.2 nm
✅ Correct: B
Second line of Balmer (n=4→n=2): \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{4}-\dfrac{1}{16}\right) = R_H\times\dfrac{3}{16}\). Comparing with first line (n=3→n=2): \(\dfrac{1}{\lambda_1} = R_H\times\dfrac{5}{36}\). \(\lambda_2 = \lambda_1\times\dfrac{5/36}{3/16} = 656\times\dfrac{20}{27} = \mathbf{486.3\text{ nm}}\).
Q5
The number of angular nodes and radial nodes in 3d orbital are respectively:
A2 and 0
B0 and 2
C1 and 2
D2 and 1
✅ Correct: A
For 3d orbital: n=3, l=2. Angular nodes \(= l = 2\). Radial nodes \(= n-l-1 = 3-2-1 = 0\). So 2 angular nodes and 0 radial nodes. Angular nodes are planar/conical surfaces; radial nodes are spherical shells.
Q6
The electronic configuration of Cu is [Ar] 3d¹⁰ 4s¹ (not [Ar] 3d⁹ 4s²) because:
A3d and 4s have same energy in Cu
BCompletely filled 3d subshell gives extra stability
C4s orbital is always filled last
DCu cannot have 9 d electrons
✅ Correct: B
Fully filled (3d¹⁰) and half-filled (3d⁵) subshells have extra stability due to symmetrical distribution of electrons and exchange energy. Cu prefers [Ar]3d¹⁰4s¹ to gain the stability of completely filled d subshell.
Q7
According to Heisenberg's uncertainty principle, if the uncertainty in the position of an electron is 10⁻¹⁰ m, the minimum uncertainty in its velocity is: (m_e = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J·s)
Which of the following sets of quantum numbers is not permissible?
An=2, l=1, m_l=0, m_s=+½
Bn=3, l=0, m_l=0, m_s=−½
Cn=4, l=3, m_l=+4, m_s=+½
Dn=2, l=1, m_l=−1, m_s=+½
✅ Correct: C
For l=3, m_l can range from −3 to +3. m_l = +4 is not allowed (exceeds l). This set is not permissible. All other options have valid quantum numbers within allowed ranges.
Q9
The ionisation energy of hydrogen atom is 13.6 eV. The energy required to ionise a He⁺ ion (Z=2) in its ground state is:
A27.2 eV
B54.4 eV
C13.6 eV
D6.8 eV
✅ Correct: B
\(E_n = -13.6\dfrac{Z^2}{n^2}\) eV. For He⁺ (Z=2, n=1): \(E_1 = -13.6\times4 = -54.4\text{ eV}\). Energy required to ionise \(= \mathbf{54.4\text{ eV}}\).
Q10
The maximum number of electrons that can be accommodated in the M shell (n=3) is:
A8
B18
C32
D2
✅ Correct: B
M shell has n=3. Subshells: 3s (2e), 3p (6e), 3d (10e). Total \(= 2+6+10 = \mathbf{18\text{ electrons}}\). General formula: max electrons in nth shell \(= 2n^2 = 2\times9 = 18\).
Q11
Which orbital has dumbbell shape along z-axis?
Ap_x
Bp_y
Cp_z
Dd_{z²}
✅ Correct: C
The p_z orbital has its electron density concentrated along the z-axis, forming a dumbbell shape with two lobes — one above and one below the xy-plane. p_x is along x-axis, p_y along y-axis.
Q12
Bohr model of atom is based on which of the following assumptions?
AElectrons can move in any orbit around nucleus
BElectrons revolve in fixed circular orbits where angular momentum is quantized as nh/2π
CElectrons continuously emit radiation as they revolve
DAll orbits have equal energy
✅ Correct: B
Bohr's second postulate: electrons move in circular orbits where angular momentum \(L = mvr = \mathbf{nh/2\pi}\) where n is a positive integer. This quantization leads to discrete energy levels and explains the hydrogen spectrum.
Q13
In hydrogen spectrum, the series that lies entirely in the ultraviolet region is:
APaschen
BBalmer
CBrackett
DLyman
✅ Correct: D
Lyman series: transitions to n=1 from higher levels. Energy is highest → UV region. Balmer series: n=2, visible region. Paschen: n=3, near IR. Brackett: n=4, IR. Pfund: n=5, far IR.
Q14
The de Broglie wavelength associated with an electron moving with velocity 10⁶ m/s is approximately: (m_e = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J·s)
The correct order of energies of orbitals in multi-electron atoms using Aufbau principle is:
A3d > 4s > 3p
B4s > 3d > 3p
C3p > 3d > 4s
D4s > 3p > 3d
✅ Correct: B
Aufbau order (filling order): 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p... So energy order for filling: \(4s < 3d\). In multi-electron atoms: 4s has lower energy than 3d and fills first. Order from low to high energy: 3p < 4s < 3d. So 4s > 3d > 3p in reverse.
Q16
Which quantum number determines the shape of an orbital?
APrincipal quantum number (n)
BAzimuthal quantum number (l)
CMagnetic quantum number (m_l)
DSpin quantum number (m_s)
✅ Correct: B
Azimuthal (angular momentum) quantum number l determines the shape of the orbital. l=0 → spherical (s); l=1 → dumbbell (p); l=2 → double dumbbell or ring (d); l=3 → complex (f). n determines size; m_l determines orientation; m_s determines spin.
Q17
The number of d electrons in Fe²⁺ (Z = 26) is:
A4
B5
C6
D8
✅ Correct: C
Fe: [Ar] 3d⁶ 4s². Fe²⁺: loses 2 electrons from 4s first. Fe²⁺: [Ar] 3d⁶. Number of 3d electrons = 6.
Q18
Photoelectric effect was explained by Einstein using the concept of:
AWave nature of light
BQuantisation of energy (photons)
CInterference of light
DDiffraction of electrons
✅ Correct: B
Einstein explained the photoelectric effect (Nobel Prize 1921) using the concept that light consists of quanta (photons), each with energy E=hν. When a photon strikes a metal surface, its energy is used to overcome the work function, and excess becomes KE of emitted electron.
Q19
The maximum number of electrons in a subshell with l = 3 (f-subshell) is:
A7
B10
C14
D6
✅ Correct: C
For l=3 (f-subshell): m_l ranges from −3 to +3 → 7 orbitals. Each orbital holds 2 electrons. Maximum electrons \(= 7\times2 = \mathbf{14}\).
Q20
Which of the following electronic configurations violates Hund's rule?
A2p: ↑↓ | _ | _
B2p: ↑ | ↑ | ↑
C2p: ↑ | ↑ | _
D2p: ↑↓ | ↑↓ | ↑↓
✅ Correct: A
Hund's rule: every orbital in a subshell is singly occupied before any is doubly occupied, and all singly occupied orbitals have the same spin. Configuration 2p: ↑↓ | _ | _ violates this rule because two electrons are paired in one orbital while other orbitals remain empty. Correct filling: ↑|↑|_ then ↑↓|↑|_.
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