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Atomic Structure Questions and Answers

21.

The wavelength of a fast moving particle (λ) is related to its momentum (p) and Planck's constant (h) as

Exam Year(s)

KEAM - 2026,

Answer: C

The relation is \(\lambda = h/p\), the de Broglie relation.

De Broglie proposed that a moving particle has a wave nature with wavelength inversely proportional to its momentum, \(\lambda = \frac{h}{p} = \frac{h}{mv}\). The faster or heavier the particle, the shorter its wavelength.

A quick unit check rules out the rest: \(h\) is in J s and \(p\) in kg m s\(^{-1}\), so \(h/p\) is in metres, a length. \(hp\), \(h^{2}/p\), \(p^{2}/h\) and \(h/p^{2}\) do not reduce to metres, so none can be a wavelength.

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22.

Consider the following statements about photoelectric effect:
(i)There is time lag between the striking of light beam and the ejection of electrons from the metal surface.
(ii)The number of electrons ejected is directly proportional to the intensity or brightness of light.
(iii)For each metal, there is a characteristic minimum frequency, known as threshold frequency (ν0) below which photoelectric effect is not observed.
(iv)At a frequency greater than ν0, the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of the wavelength of the light used.

Which of the above statements are true?

Exam Year(s)

KEAM - 2026,

Answer: C

The true statements are (ii) and (iii).

(ii) is true: above the threshold, each photon ejects at most one electron, and a brighter beam carries more photons, so the number of electrons is proportional to the intensity. (iii) is true: every metal has a threshold frequency ν0, below which no electrons are emitted however bright the light.

(i) is false: emission is practically instantaneous, with no measurable time lag, one of the observations that classical wave theory could not explain. (iv) is false: the kinetic energy increases with the frequency of the light, so it decreases as the wavelength increases.

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23.

According to Bohr's theory the radius of the second stationary state of hydrogen atom is

Exam Year(s)

KEAM - 2026,

Answer: E

The radius is 211.6 pm.

In Bohr's model, \(r_{n} = 52.9\,n^{2}\) pm for hydrogen (Z = 1).

\[ r_{2} = 52.9 \times 2^{2} = 211.6\ \text{pm} \]

105.8 pm uses n instead of \(n^{2}\), and 423.2 pm uses \(n^{3}\) (equivalently, n = 2 with an extra factor 2). The radius grows as the square of n.

The general Bohr result is \(r_{n} = 52.9\frac{n^{2}}{Z}\) pm, where 52.9 pm is the Bohr radius \(a_{0}\) of the ground state. 21.16 pm and 5.29 pm are smaller than the ground-state radius itself, which no excited state can be.

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24.

The product of uncertainty in position and uncertainty in momentum is constant. This is known as

Exam Year(s)

KEAM - 2026,

Answer: C

This is Heisenberg's principle, the uncertainty principle.

It states that the position and momentum of a particle such as an electron cannot both be known exactly at the same time: \(\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\), so the product of the two uncertainties has a fixed lower limit.

Hund's rule is about filling degenerate orbitals singly before pairing, and the Aufbau principle is about filling orbitals in order of increasing energy. Bohr's correspondence principle links quantum and classical results at large quantum numbers, and no 'Einstein's principle' describes this relationship.

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25.

For a given sub-shell with azimuthal quantum number 'l' the number of magnetic quantum numbers possible is

Exam Year(s)

KEAM - 2026,

Answer: E

The number of magnetic quantum numbers is \(2l + 1\).

For a given \(l\), the magnetic quantum number \(m_l\) takes every whole value from \(-l\) through 0 to \(+l\). That is \(l\) negative values, \(l\) positive values and zero, so \(2l + 1\) values in all, one for each orbital in the sub-shell.

Check: a p sub-shell (\(l = 1\)) has \(m_l = -1, 0, +1\), three orbitals, and a d sub-shell (\(l = 2\)) has five. \(l(l+1)\), \(l+1\), \(2(l+1)\) and \(l-1\) give 2, 2, 4 and 0 for \(l = 1\), none of which is right.

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26.

The number of nodal surfaces present in '4s' orbital is

Exam Year(s)

KEAM - 2026,

Answer: A

A 4s orbital has three nodal surfaces.

The total number of nodes in an orbital is \(n - 1\), made up of \(l\) angular nodes and \(n - l - 1\) radial (spherical) nodes. For 4s, \(n = 4\) and \(l = 0\):

\[ \begin{aligned} \text{radial nodes} &= n - l - 1 = 4 - 0 - 1 = 3 \\ \text{angular nodes} &= l = 0 \end{aligned} \]

So all three nodal surfaces are spheres. Four is the value of \(n\), not the node count, and one or two would belong to the 2s and 3s orbitals.

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27.

The minimum energy required to remove an electron from sodium atom is 3.313 × 10-19 J. What is the maximum wavelength of radiation that will eject photoelectron from sodium metal? [h = 6.626 × 10-34 Js]

Exam Year(s)

KEAM - 2026,

Answer: E

The maximum wavelength is 600 nm.

The longest wavelength that can eject an electron carries exactly the minimum energy, \(E = \frac{hc}{\lambda}\), with \(c = 3 \times 10^{8}\) m s-1:

\[ \begin{aligned} \lambda &= \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3.313 \times 10^{-19}} \\ &= 6 \times 10^{-7}\ \text{m} = 600\ \text{nm} \end{aligned} \]

Since \(6.626 = 2 \times 3.313\), the numbers simplify neatly to \(2 \times 3 \times 10^{-7}\) m. Shorter wavelengths such as 400 or 500 nm would also eject electrons, but they are not the maximum; 700 nm photons carry too little energy.

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28.

Which of the following statements are correct about the postulates of quantum mechanical model of an atom?

(i) The energy of electron in atom is quantized.
(ii) The existence of quantized electronic energy level is a result of the particle property of electrons.
(iii) The path of the electron can be determined accurately.
(iv) In a multi electron atom, the electrons are filled in various orbitals in the order of increasing energy.
(v) The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function.

Exam Year(s)

KEAM - 2026,

Answer: A

The correct statements are (i), (iv) and (v), so the answer is (i), (iv), (v).

In the quantum mechanical model, the energy of an electron in an atom is quantized (i); electrons fill orbitals in order of increasing energy (iv, the aufbau principle); and the probability of finding an electron at a point is proportional to \(|\psi|^{2}\), the square of the orbital wave function (v).

Statement (ii) is false: quantized energy levels arise from the wave nature of the electron, not its particle nature. Statement (iii) is false: by the uncertainty principle the path of an electron cannot be determined exactly. Every other option includes (ii) or (iii).

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29.

Which of the following d-orbital does not have four lobes?

Exam Year(s)

KEAM - 2026,

Answer: E

The orbital without four lobes is \(\mathrm{d_{z^{2}}}\).

The \(\mathrm{d_{z^{2}}}\) orbital has a different shape from the other four d orbitals: two lobes along the z-axis with a doughnut-shaped ring of electron density (a collar) around the middle in the xy-plane.

The other four, \(\mathrm{d_{xy}}\), \(\mathrm{d_{yz}}\), \(\mathrm{d_{xz}}\) and \(\mathrm{d_{x^{2}-y^{2}}}\), each have four lobes in a cloverleaf pattern. The first three have their lobes between the axes, and \(\mathrm{d_{x^{2}-y^{2}}}\) has its lobes along the x and y axes.

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Answer: B

The incorrect statement is For 1s orbital the probability density is maximum at the nucleus and it increases sharply as we move away from it.

For a 1s electron the probability density is greatest at the nucleus and then decreases sharply as the distance from the nucleus increases. The statement gets the direction wrong.

The others are correct: the square of the wave function gives the probability density; for 2s the density falls to zero at a radial node and then rises again; a node is where the probability density is zero; and an ns orbital has n - 1 radial nodes, so 3s has 2.

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