Quantum Mechanical Model of an Atom - Free MCQ Practice Test with solutions,

For 1s-orbital radial wave function (Ft) is maximum at r - 0, and falls rapidly as r increases thus, (a) correct.
For2s-orbital, radial wave function (R) is maximum at (r = 0), falls to zero and further decreases with r. There appears radial nodes. Thus (b) correct.
For2p-orbital, radial wave function is zero at r = 0, reaches maximum value (at r = a₀) and then falls thus (c) is correct.

Correct Answer : b

Explanation : (a) It represents R2 vs r for 1s

(b) It represents R2 vs r for2s

(c) It represents R2 vs r for 2p

(a) Describes radial wave function as a function of r for 1s
(b) Describes radial probability function as a function of r for 2s
(c) Describes radial wave function as a function r for 2s
(d) Describes radial probability function as a function of r for 2p

in case of 1s, falls as r increases thus, (A) is for 1s.
In case of 2s, is maximum in the vicinity o f the nucleus, falls to zero giving spherical nodes and then rises to second highest value. Thus, (6) is for2s

 Thus, is d e pendent on r true

(b) and | | are dependent on r, θ and  : true
(c) Angular wave function is determined by /land m, and not by n: true

3s is spherically symmetrical and its electron density is maximum at the nucleus. It decreases with r.

H = T + V

The number of angular nodes in an orbital is equal to the value of the azimuthal quantum number (l). For the 4s orbital, l = 0, so the number of angular nodes is 0.
The number of radial nodes is calculated using the formula: radial nodes = n – l – 1.
For 3d: n = 3, l = 2, so radial nodes = 3 – 2 – 1 = 0.
For 2p: n = 2, l = 1, so radial nodes = 2 – 1 – 1 = 0.
Thus, both 3d and 2p orbitals have 0 radial nodes, which equals the number of angular nodes in the 4s orbital.
Therefore, the correct answer is Both (A) and (B).

By Born’s interpretation of the wave function, the square of the magnitude of the wave function gives probability density:

∣ψ∣² = probability density

So ∣ψ∣²dV is the probability of finding the particle in the small volume element dV around that point.

Hence option (a) is the correct answer.

For n nodes in a string

The wave function describ es properties of the orbital and the electron that occupies the orbital
(a) Type of orbital is described : true
(b) Energy of electron : true
(c) Shape of the orbital: true
(d) Probability: true

For2s-electron of H-atom (first excited state)

For 1s-electron of H-atom in ground state

At point Ar = a₀ r at point Br > a₀
(a) True (b) True (c) True (d) True

Radial probability increases as r increase reaches a maximum value when r = a0 (Bohr’s radius) and then falls. When radial probability is very small.
Thus, (a) and (c) are true.