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Atomic Models, Dual Nature of Matter and Uncertainty Principle — NEET Chemistry MCQs with Solutions

Free NEET Chemistry Atomic Models, Dual Nature of Matter and Uncertainty Principle MCQs with step-by-step solutions (121 questions). Part of Atomic Structure. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
According to Bohr's model of hydrogen atom, which of the following statements is incorrect?
A. Radius of the 6th orbit is three times larger than that of the 4th orbit.  ✓ Correct
B. Radius of the 3rd orbit is nine times larger than that of the 1st orbit.
C. Radius of the 8th orbit is four times larger than that of the 4th orbit.
D. Radius of the 4th orbit is four times larger than that of the 2nd orbit.
Solution: In the Bohr model $r_n \propto n^2$. So $\frac{r_6}{r_4} = \frac{36}{16} = 2.25$, not 3 — statement (1) is incorrect. The others check out: $\frac{r_3}{r_1} = 9$, $\frac{r_8}{r_4} = \frac{64}{16} = 4$ and $\frac{r_4}{r_2} = \frac{16}{4} = 4$.
Q2 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
Which of the following postulates of Bohr's model of hydrogen atom is not in agreement with the quantum mechanical model of an atom?
A. When an electron makes a transition from a higher energy stationary state to a lower energy stationary state, it emits a photon of light.
B. An atom can take only certain distinct energies $E_1$, $E_2$, $E_3$, etc. These allowed states of constant energy are called the stationary states of the atom.
C. The electron in a hydrogen atom's stationary state moves in a circle around the nucleus.  ✓ Correct
D. An atom in a stationary state does not emit electromagnetic radiation as long as it stays in the same state.
Solution: The quantum mechanical model rejects definite trajectories: by the Heisenberg uncertainty principle an electron cannot follow a well-defined circular path; only probability distributions (orbitals) are meaningful. Quantized energies and photon emission on transitions are retained in the quantum model.
Q3 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
Which of the following statements are correct, if the threshold frequency of caesium is $5.16 \times 10^{14}$ Hz? A. When Cs is placed inside a vacuum chamber with an ammeter connected to it and yellow light is focused on Cs, the ammeter shows the presence of current. B. When the brightness of the yellow light is dimmed, the value of the current in the ammeter is reduced. C. When a red light is used instead of the yellow light, the current produced is higher with respect to the yellow light. D. When a blue light is used, the ammeter shows the formation of current. E. When a white light is used, the ammeter shows formation of current. Choose the correct answer from the options given below:
A. A, C, D and E only
B. A, B, D and E only  ✓ Correct
C. B, C and D only
D. A, D and E only
Solution: Yellow light ($\lambda \approx 580$ nm, $\nu \approx 5.2 \times 10^{14}$ Hz) is just above the threshold, so it ejects photoelectrons (A). Photocurrent is proportional to intensity, so dimming reduces it (B). Red light ($\nu \approx 4.3 \times 10^{14}$ Hz) is below threshold — no current at all, so C is wrong. Blue light is above threshold (D), and white light contains blue/violet components (E).
Q4 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The radius of the first excited state of the helium ion (He⁺) is given as ($a_0$ = radius of the first stationary state of hydrogen atom):
A. $r = 2a_0$  ✓ Correct
B. $r = 4a_0$
C. $r = \frac{a_0}{2}$
D. $r = \frac{a_0}{4}$
Solution: For a hydrogen-like ion $r_n = \frac{n^2 a_0}{Z}$. The first excited state means $n = 2$, and for He⁺, $Z = 2$: $r = \frac{4a_0}{2} = 2a_0$.
Q5 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this? Given: Rydberg constant $R_H = 10^5$ cm⁻¹, $h = 6.6 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s
A. Lyman series, $\infty \rightarrow 1$
B. Balmer series, $\infty \rightarrow 2$
C. Paschen series, $5 \rightarrow 3$
D. Paschen series, $\infty \rightarrow 3$  ✓ Correct
Solution: $\bar{\nu} = \frac{1}{\lambda} = \frac{1}{900 \times 10^{-7}\text{ cm}} \approx 1.11 \times 10^4$ cm⁻¹. Then $\frac{\bar{\nu}}{R_H} = \frac{1.11 \times 10^4}{10^5} = 0.111 = \frac{1}{9} = \frac{1}{3^2} - \frac{1}{\infty^2}$. So the transition is $\infty \rightarrow 3$: the series limit of the Paschen series.
Q6 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 'A' $\times 10^{12}$ hertz and that has a radiant intensity in that direction of 'B' watt per steradian. 'A' and 'B' are respectively:
A. 450 and 683
B. 450 and $\frac{1}{683}$
C. 540 and $\frac{1}{683}$  ✓ Correct
D. 540 and 683
Solution: By the SI definition, the candela is the luminous intensity of a source emitting monochromatic radiation of frequency $540 \times 10^{12}$ Hz (green light, ~555 nm) with a radiant intensity of $\frac{1}{683}$ watt per steradian.
Q7 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
According to Bohr's model, the highest kinetic energy is associated with the electron in the
A. first orbit of H atom
B. first orbit of He⁺  ✓ Correct
C. second orbit of He⁺
D. second orbit of Li²⁺
Solution: Kinetic energy in a Bohr orbit is $KE = 13.6\,\frac{Z^2}{n^2}$ eV. Comparing $\frac{Z^2}{n^2}$: H ($n=1$): 1; He⁺ ($n=1$): 4; He⁺ ($n=2$): 1; Li²⁺ ($n=2$): $\frac{9}{4} = 2.25$. The first orbit of He⁺ has the highest value.
Q8 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
If the radius of the first orbit of hydrogen atom is $a_0$, then de Broglie's wavelength of the electron in the 3rd orbit is
A. $6\pi a_0$  ✓ Correct
B. $\frac{a_0}{6\pi}$
C. $\frac{a_0}{3\pi}$
D. $3\pi a_0$
Solution: Bohr quantization gives $2\pi r_n = n\lambda$, so $\lambda = \frac{2\pi r_n}{n}$. With $r_n = n^2 a_0$, $\lambda = 2\pi n a_0$. For $n = 3$: $\lambda = 6\pi a_0$.
Q9 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The energy of an electron in the first Bohr orbit of hydrogen atom is $-2.18 \times 10^{-18}$ J. Its energy in the third Bohr orbit is ______.
A. $\frac{1}{27}$ of this value
B. $\frac{1}{9}$th of this value  ✓ Correct
C. one third of this value
D. three times this value
Solution: $E_n = \frac{E_1}{n^2}$. For $n = 3$: $E_3 = \frac{E_1}{9}$, i.e. one-ninth of the first-orbit energy ($-2.42 \times 10^{-19}$ J).
Q10 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The radius of the 2nd orbit of Li²⁺ is $x$. The expected radius of the 3rd orbit of Be³⁺ is
A. $\frac{16x}{27}$
B. $\frac{4x}{9}$
C. $\frac{9x}{4}$
D. $\frac{27x}{16}$  ✓ Correct
Solution: $r_n = \frac{n^2 a_0}{Z}$. For Li²⁺ ($Z=3$, $n=2$): $x = \frac{4a_0}{3}$, so $a_0 = \frac{3x}{4}$. For Be³⁺ ($Z=4$, $n=3$): $r = \frac{9a_0}{4} = \frac{9}{4} \cdot \frac{3x}{4} = \frac{27x}{16}$.
Q11 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The energy of one mole of photons of radiation of wavelength 300 nm is (Given: $h = 6.63 \times 10^{-34}$ J s, $N_A = 6.02 \times 10^{23}$ mol⁻¹, $c = 3 \times 10^8$ m s⁻¹)
A. 235 kJ mol⁻¹
B. 325 kJ mol⁻¹
C. 399 kJ mol⁻¹  ✓ Correct
D. 435 kJ mol⁻¹
Solution: Energy per photon $= \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.63 \times 10^{-19}$ J. Per mole: $6.63 \times 10^{-19} \times 6.02 \times 10^{23} \approx 3.99 \times 10^5$ J = 399 kJ mol⁻¹.
Q12 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: threshold frequency of platinum $= 1.3 \times 10^{15}$ s⁻¹ and $h = 6.6 \times 10^{-34}$ J s]
A. $3.21 \times 10^{-14}$ J
B. $6.24 \times 10^{-16}$ J
C. $8.58 \times 10^{-19}$ J  ✓ Correct
D. $9.76 \times 10^{-20}$ J
Solution: Minimum photon energy = work function $= h\nu_0 = 6.6 \times 10^{-34} \times 1.3 \times 10^{15} = 8.58 \times 10^{-19}$ J.
Q13 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
If the radius of the 3rd Bohr orbit of hydrogen atom is $r_3$ and the radius of the 4th Bohr orbit is $r_4$, then
A. $r_4 = \frac{9}{16} r_3$
B. $r_4 = \frac{16}{9} r_3$  ✓ Correct
C. $r_4 = \frac{3}{4} r_3$
D. $r_4 = \frac{4}{3} r_3$
Solution: $r_n \propto n^2$, so $\frac{r_4}{r_3} = \frac{4^2}{3^2} = \frac{16}{9}$, i.e. $r_4 = \frac{16}{9} r_3$.
Q14 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
If the Thomson model of the atom were correct, then the result of Rutherford's gold foil experiment would have been:
A. α-particles pass through the gold foil deflected by small angles and with reduced speed  ✓ Correct
B. α-particles are deflected over a wide range of angles
C. all α-particles get bounced back by 180°
D. all of the α-particles pass through the gold foil without decrease in speed
Solution: In Thomson's plum-pudding model the positive charge is spread uniformly over the whole atom, so the electric field anywhere is weak. α-particles would then suffer only small deflections (and lose a little energy to the distributed charge) — never the large-angle scattering Rutherford actually observed.
Q15 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is
A. 2 and 1  ✓ Correct
B. 2 and 2
C. 3 and 1
D. 1 and 1
Solution: The radioactive isotope of hydrogen is tritium, ³H (mass number 3, atomic number 1). It has $3 - 1 = 2$ neutrons and 1 electron.
Q16 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · hard
According to Bohr's atomic theory: (A) Kinetic energy of the electron is $\propto \frac{Z^2}{n^2}$ (B) The product of velocity ($v$) of the electron and principal quantum number ($n$), $vn \propto Z^2$ (C) Frequency of revolution of the electron in an orbit is $\propto \frac{Z^3}{n^3}$ (D) Coulombic force of attraction on the electron is $\propto \frac{Z^3}{n^4}$ Choose the most appropriate answer from the options given below:
A. A and D only  ✓ Correct
B. A only
C. C only
D. A, C and D only
Solution: (A) $KE = 13.6\frac{Z^2}{n^2}$ eV — correct. (B) $v \propto \frac{Z}{n}$, so $vn \propto Z$, not $Z^2$ — wrong. (C) Frequency $f = \frac{v}{2\pi r} \propto \frac{Z/n}{n^2/Z} = \frac{Z^2}{n^3}$, not $\frac{Z^3}{n^3}$ — wrong. (D) $F = \frac{kZe^2}{r^2} \propto \frac{Z}{(n^2/Z)^2} = \frac{Z^3}{n^4}$ — correct. Hence A and D only.
Q17 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The region in the electromagnetic spectrum where the Balmer series lines appear is
A. microwave
B. ultraviolet
C. visible  ✓ Correct
D. infrared
Solution: Balmer series lines (transitions to $n = 2$) lie in the visible region (656 nm to 365 nm). Lyman is UV; Paschen, Brackett and Pfund are infrared.
Q18 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The shortest wavelength of H-atom in the Lyman series is $\lambda_1$. The longest wavelength in the Balmer series of He⁺ is
A. $\frac{9\lambda_1}{5}$  ✓ Correct
B. $\frac{5\lambda_1}{9}$
C. $\frac{27\lambda_1}{5}$
D. $\frac{36\lambda_1}{5}$
Solution: Shortest Lyman line of H: $\frac{1}{\lambda_1} = R(1 - 0) = R$, so $\lambda_1 = \frac{1}{R}$. Longest Balmer line of He⁺ ($Z = 2$, $3 \rightarrow 2$): $\frac{1}{\lambda} = RZ^2\left(\frac{1}{4} - \frac{1}{9}\right) = 4R \cdot \frac{5}{36} = \frac{5R}{9}$. Hence $\lambda = \frac{9}{5R} = \frac{9\lambda_1}{5}$.
Q19 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The difference between the radii of the 3rd and 4th orbits of Li²⁺ is $\Delta R_1$. The difference between the radii of the 3rd and 4th orbits of He⁺ is $\Delta R_2$. The ratio $\Delta R_1 : \Delta R_2$ is
A. 8 : 3
B. 3 : 8
C. 2 : 3  ✓ Correct
D. 3 : 2
Solution: $r_n = \frac{n^2 a_0}{Z}$, so $\Delta R = \frac{(16 - 9)a_0}{Z} = \frac{7a_0}{Z}$. For Li²⁺ ($Z = 3$): $\Delta R_1 = \frac{7a_0}{3}$; for He⁺ ($Z = 2$): $\Delta R_2 = \frac{7a_0}{2}$. Ratio $= \frac{1/3}{1/2} = 2 : 3$.
Q20 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
For the Balmer series in the spectrum of the H atom, $\bar{\nu} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, the correct statements among (I) to (IV) are: I. As wavelength decreases, the lines in the series converge. II. The integer $n_1$ is equal to 2. III. The line of longest wavelength corresponds to $n_2 = 3$. IV. The ionization energy of hydrogen can be calculated from the wave number of these lines.
A. II, III, IV
B. I, III, IV
C. I, II, III  ✓ Correct
D. I, II, IV
Solution: For Balmer, $n_1 = 2$ (II correct). The smallest energy gap, $3 \rightarrow 2$, gives the longest wavelength (III correct). Successive lines crowd together towards the series limit as wavelength decreases (I correct). The Balmer series limit gives the energy to remove the electron from $n = 2$, not from the ground state, so the ionization energy of hydrogen cannot be obtained from these lines (IV wrong).
Q21 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The radius of the second Bohr orbit, in terms of the Bohr radius $a_0$, in Li²⁺ is
A. $\frac{2a_0}{3}$
B. $\frac{4a_0}{3}$  ✓ Correct
C. $\frac{4a_0}{9}$
D. $\frac{2a_0}{9}$
Solution: $r_n = \frac{n^2 a_0}{Z}$. For Li²⁺ ($Z = 3$) with $n = 2$: $r = \frac{4a_0}{3}$.
Q22 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The de Broglie wavelength of an electron in the 4th Bohr orbit is ($a_0$ = Bohr radius)
A. $6\pi a_0$
B. $2\pi a_0$
C. $8\pi a_0$  ✓ Correct
D. $4\pi a_0$
Solution: $2\pi r_n = n\lambda \Rightarrow \lambda = \frac{2\pi r_n}{n} = \frac{2\pi n^2 a_0}{n} = 2\pi n a_0$. For $n = 4$: $\lambda = 8\pi a_0$.
Q23 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
If $p$ is the momentum of the fastest electron ejected from a metal surface after irradiation with light of wavelength $\lambda$, then for a photoelectron of momentum $1.5p$ the wavelength of the light should be (Assume the kinetic energy of the ejected photoelectron to be very high compared to the work function.)
A. $\frac{2}{3}\lambda$
B. $\frac{4}{9}\lambda$  ✓ Correct
C. $\frac{3}{4}\lambda$
D. $\frac{1}{2}\lambda$
Solution: With the work function negligible, $\frac{hc}{\lambda} \approx KE = \frac{p^2}{2m}$, so $\lambda \propto \frac{1}{p^2}$. For momentum $1.5p$: $\lambda' = \frac{\lambda}{(1.5)^2} = \frac{\lambda}{2.25} = \frac{4}{9}\lambda$.
Q24 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · hard
For any given series of spectral lines of atomic hydrogen, let $\Delta\bar{\nu} = \bar{\nu}_{max} - \bar{\nu}_{min}$ be the difference in maximum and minimum wave numbers in cm⁻¹. The ratio $\Delta\bar{\nu}_{Lyman} : \Delta\bar{\nu}_{Balmer}$ is
A. 4 : 1
B. 27 : 5
C. 9 : 4  ✓ Correct
D. 5 : 4
Solution: Lyman: $\bar{\nu}_{max} = R(1 - 0) = R$, $\bar{\nu}_{min} = R(1 - \frac{1}{4}) = \frac{3R}{4}$, so $\Delta\bar{\nu} = \frac{R}{4}$. Balmer: $\bar{\nu}_{max} = \frac{R}{4}$, $\bar{\nu}_{min} = R(\frac{1}{4} - \frac{1}{9}) = \frac{5R}{36}$, so $\Delta\bar{\nu} = \frac{9R - 5R}{36} = \frac{R}{9}$. Ratio $= \frac{1/4}{1/9} = 9 : 4$.
Q25 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
Which one of the following statements about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by $a_0$.)
A. The electron can be found at a distance $2a_0$ from the nucleus.
B. The magnitude of the potential energy is double that of its kinetic energy on an average.
C. The probability density of finding the electron is maximum at the nucleus.
D. The total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus.  ✓ Correct
Solution: The total energy of a 1s electron is a constant ($-13.6$ eV) — it does not depend on the instantaneous distance from the nucleus, so statement (4) is incorrect. The 1s probability density $|\psi|^2$ is indeed maximum at the nucleus, the electron has nonzero probability at $2a_0$, and the virial theorem gives $|PE| = 2\,KE$.
Q26 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The ratio of the shortest wavelengths of two spectral series of the hydrogen spectrum is found to be about 9. The spectral series are:
A. Paschen and Pfund
B. Lyman and Paschen  ✓ Correct
C. Balmer and Brackett
D. Brackett and Pfund
Solution: The shortest wavelength (series limit) of a series ending at $n_1$ is $\lambda_{min} = \frac{n_1^2}{R}$, so the ratio of series limits is $\frac{n_1^2}{n_1'^2}$. A ratio of 9 needs $\frac{n_1'}{n_1} = 3$: Paschen ($n_1 = 3$) and Lyman ($n_1 = 1$), giving $\frac{9}{1} = 9$.
Q27 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
For emission lines of atomic hydrogen from $n_i = 8$ to $n_f = n$, the plot of wave number ($\bar{\nu}$) against $\frac{1}{n^2}$ will be (The Rydberg constant $R_H$ is in wave number units.)
A. linear with intercept $-R_H$
B. non-linear
C. linear with slope $R_H$  ✓ Correct
D. linear with slope $-R_H$
Solution: $\bar{\nu} = R_H\left(\frac{1}{n^2} - \frac{1}{8^2}\right) = R_H \cdot \frac{1}{n^2} - \frac{R_H}{64}$. Plotted against $\frac{1}{n^2}$ this is a straight line of slope $+R_H$ (with intercept $-\frac{R_H}{64}$, not $-R_H$).
Q28 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · easy
The ground state energy of hydrogen atom is $-13.6$ eV. The energy of the second excited state of the He⁺ ion in eV is
A. $-3.4$
B. $-54.4$
C. $-27.2$
D. $-6.04$  ✓ Correct
Solution: $E_n = -13.6\,\frac{Z^2}{n^2}$ eV. Second excited state means $n = 3$; for He⁺, $Z = 2$: $E = -13.6 \times \frac{4}{9} = -6.04$ eV.
Q29 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
The de Broglie wavelength ($\lambda$) associated with a photoelectron varies with the frequency ($\nu$) of the incident radiation as ($\nu_0$ is the threshold frequency):
A. $\lambda \propto \frac{1}{(\nu - \nu_0)}$
B. $\lambda \propto \frac{1}{(\nu - \nu_0)^{1/2}}$  ✓ Correct
C. $\lambda \propto \frac{1}{(\nu - \nu_0)^{3/2}}$
D. $\lambda \propto \frac{1}{(\nu - \nu_0)^{1/4}}$
Solution: The photoelectron's kinetic energy is $\frac{p^2}{2m} = h(\nu - \nu_0)$, so $p = \sqrt{2mh(\nu - \nu_0)}$. Then $\lambda = \frac{h}{p} \propto \frac{1}{(\nu - \nu_0)^{1/2}}$.
Q30 — Atomic Models, Dual Nature of Matter and Uncertainty Principle · medium
What is the work function of the metal if light of wavelength 4000 Å generates photoelectrons of velocity $6 \times 10^5$ m s⁻¹ from it? (Mass of electron $= 9 \times 10^{-31}$ kg, velocity of light $= 3 \times 10^8$ m s⁻¹, Planck's constant $= 6.626 \times 10^{-34}$ J s, charge of electron $= 1.6 \times 10^{-19}$ J eV⁻¹)
A. 4.0 eV
B. 2.1 eV  ✓ Correct
C. 0.9 eV
D. 3.1 eV
Solution: Photon energy $= \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = 4.97 \times 10^{-19}$ J. Kinetic energy $= \frac{1}{2}mv^2 = \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2 = 1.62 \times 10^{-19}$ J. Work function $= (4.97 - 1.62) \times 10^{-19} = 3.35 \times 10^{-19}$ J $= \frac{3.35 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.1$ eV.