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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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451 Medium
451 of 492
Application +4 | -1
The energy ∆E corresponding to intense yellow line of sodium of λ,  589  nm\lambda ,\;589\;\mathrm{nm} is 
View answer and explanation
Correct option: A
2.10 eV
Option A (2.10 eV) is correct because it represents the energy calculated using the formula E=hc/λE = hc/\lambda and converting Joules to electron volts.
452 Medium
452 of 492
Remembering +4 | -1
Millikan’s oil drop experiment is used to find:
View answer and explanation
Correct option: B
Electronic charge
Option B (Electronic charge) is correct because Millikan's oil drop experiment was designed to measure the elementary electric charge (e).
453 Easy
453 of 492
Understanding +4 | -1
An electron from one Bohr stationary orbit can go to next higher orbit
View answer and explanation
Correct option: C
By absorption of electromagnetic radiation of particular frequency
Option C (By absorption of electromagnetic radiation of particular frequency) is correct because an electron transitions to a higher energy level by absorbing a photon with energy equal to the energy difference between the orbits.
454 Medium
454 of 492
Application +4 | -1
Which one of the following groupings represents a collection of isoelectronic species? (At. no. Cs=55, Br=35)
View answer and explanation
Correct option: B
N3−,F−,Na+N ^{3-}, F ^{-}, Na ^{+}
Option B (N3−,F−,Na+N ^{3-}, F ^{-}, Na ^{+}) is correct because each of these ions has 10 electrons: N3- (7+3=10), F- (9+1=10), and Na+ (11-1=10).
455 Easy
455 of 492
Remembering +4 | -1
Particle having mass 200 times that of an electron is:
View answer and explanation
Correct option: C
Meson
Option C is correct because a meson has a mass approximately 200 times that of an electron.
456 Easy
456 of 492
Application +4 | -1
The species having more electrons than neutrons is:
View answer and explanation
Correct option: C
O2−{O^{2 - }}
Option C (O2−{O^{2 - }}) is correct because it has 10 electrons and 8 neutrons, thus having more electrons than neutrons.
457 Easy
457 of 492
Remembering +4 | -1
Photoelectric effect was discovered by :
View answer and explanation
Correct option: D
Hertz
Option D (Hertz) is correct because Heinrich Hertz first observed the photoelectric effect in 1887.
458 Medium
458 of 492
Analyzing +4 | -1
Which of the following options is incorrect regarding Bohr’s model of an atom
View answer and explanation
Correct option: C
Ratio of magnitude of KE to that of PE of first Bohr orbit in ‘H’ will be in the ratio 1:2
Option D is incorrect because the speed of an electron in the nth shell decreases as n increases, so the order should be reversed.
459 Hard
459 of 492
Understanding +4 | -1
The ratio of specific charges of α\alpha-particle and deuteron is
View answer and explanation
Correct option: C
1:11: 1
Option C (1:1) is correct because the specific charge (charge/mass) ratio for an alpha particle (2He4) is (2e/4m) and for a deuteron (1H2) is (e/2m), making the ratio 1:1.
460 Medium
460 of 492
Analyzing +4 | -1
Statement - I: For Hydrogen spectrum all Balmer lines in the emission spectrum have longer
wavelengths than that of any line of Lyman series.

Statement - II: Transition energy is always greater when electrons come to an orbit with
lower orbit number than an electron come to an orbit with higher orbit number, irrespective to
higher orbit from which electron is coming.
View answer and explanation
Correct option: A
If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I
Option A (If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I) is correct because the transition energy determines the wavelength, and transitions to lower orbits have higher energy and thus shorter wavelengths.
461 Hard
461 of 492
Application +4 | -1
In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit n=1 to the period of revolution of the electron in the orbit n=2 is:
View answer and explanation
Correct option: D
1∶8
Option D (1∶8) is correct because the period of revolution is proportional to n3n^3, so the ratio of periods is (1/2)3=1/8(1/2)^3 = 1/8.
462 Medium
462 of 492
Remembering +4 | -1
α\alpha-particles can be detected by using
View answer and explanation
Correct option: C
Zinc sulphide screen
Option C (Zinc sulphide screen) is correct because alpha particles cause scintillations (small flashes of light) when they strike a zinc sulphide screen, making them detectable.
463 Hard
463 of 492
Application +4 | -1
Density of the electron is:
View answer and explanation
Correct option: A
2.77×1012  g/mL2.77 \times {10^{12}}\;\mathrm{g}/\mathrm{mL}
Option A (2.77×1012  g/mL2.77 \times {10^{12}}\;\mathrm{g}/\mathrm{mL}) is correct because it represents the calculated density of an electron based on its mass and estimated volume.
464 Medium
464 of 492
Understanding +4 | -1
The statements, which is/are correct:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all three statements regarding the number of nodes (total, radial, and angular) in an orbital are accurate.
465 Hard
465 of 492
Application +4 | -1
What accelerating potential is needed to produce an electron beam with an effective wavelength of 0.0900.090?
View answer and explanation
Correct option: A
1.86×104eV1.86 \times {10^4}\mathrm{eV}
Option A (1.86×104eV1.86 \times {10^4}\mathrm{eV}) is correct because it is the accelerating potential required to produce the specified electron beam wavelength, calculated using de Broglie wavelength and kinetic energy relations.
466 Easy
466 of 492
Application +4 | -1
Which one of the following ions is not isoelectronic with O2−?{\mathrm{O}^{2 - }}?
View answer and explanation
Correct option: A
Ti+T{i^ + }
Option A (Ti+T{i^ + }) is correct because it has 21 electrons, while O2−{\mathrm{O}^{2 - }} has 10 electrons. Isoelectronic species have the same number of electrons.
467 Easy
467 of 492
Remembering +4 | -1
The principal quantum number n can have integral values ranging from:
View answer and explanation
Correct option: B
1  to  ∞1\mathrm{\;to\;}\infty
Option B is correct because the principal quantum number n can have integer values starting from 1 and extending to infinity.
468 Easy
468 of 492
Remembering +4 | -1
Isotopes have
View answer and explanation
Correct option: D
Different mass number
Option D (Different mass number) is correct because isotopes of an element have the same number of protons but different numbers of neutrons, leading to different mass numbers.
469 Easy
469 of 492
Analyzing +4 | -1
The electronic configuration 1s2,2s22p6,3s13p11{s^2},2{s^2}2{p^6},3{s^1}3{p^1} correctly describes:
View answer and explanation
Correct option: C
Excited state of Mg
Option C (Excited state of Mg) is correct because the ground state configuration of Mg is 1s22s22p63s21s^22s^22p^63s^2, and the given configuration shows one electron promoted from 3s to 3p.
470 Medium
470 of 492
Application +4 | -1
The frequency of first line of Balmer series in hydrogen atom is v0{v_0}. The frequency of corresponding line emitted by singly ionised helium atom is :
View answer and explanation
Correct option: B
4v04{v_0}
The correct answer is B (4v04{v_0}) because the frequency of the spectral line is proportional to the square of the atomic number (Z2). For He+, Z=2, so the frequency is 4 times that of hydrogen.
471 Easy
471 of 492
Application +4 | -1
The energy of an electron in Bohr's orbit of hydrogen atom is −13.6 eV-13.6 \mathrm{~eV}. The total electronic energy of a hypothetical He\mathrm{He} atom in which there are no electron-electron repulsions or interactions is
View answer and explanation
Correct option: C
−108.8 eV-108.8 \mathrm{~eV}
The correct answer is C (-108.8 eV) because the total electronic energy is calculated by multiplying the energy of a hydrogen-like atom by Z2Z^2 and then by the number of electrons.
472 Hard
472 of 492
Application +4 | -1
I2{I_2} molecule dissociates into atoms after absorbing light of 4500 A°. If one quantum of energy is absorbed by each molecule, the KE of iodine atoms will be (BE of I2=240  kJ/mol{\mathrm{I}_2} = 240\,\,kJ/mol )
View answer and explanation
Correct option: B
0.216×10−19J0.216 \times {10^{ - 19}}J
Option B (0.216×10−19J0.216 \times {10^{ - 19}}J) is correct because it represents the kinetic energy of the iodine atoms calculated by subtracting the bond energy from the energy of the absorbed photon.
473 Easy
473 of 492
Application +4 | -1
The wavelength of shortest wavelength transition in Lyman series of atomic Hydrogen is
View answer and explanation
Correct option: D
911.9 A∘911.9\,{A^ \circ }
Option D (911.9 A∘911.9\,{A^ \circ }) is correct because it represents the shortest wavelength (highest energy) transition in the Lyman series, corresponding to the transition from infinity to the ground state (n=1).
474 Hard
474 of 492
Understanding +4 | -1
The first orbital of H is represented by :ψ=1π(1a0)3/2e−r/a0,\mathrm{\psi } = \frac{1}{{\sqrt \mathrm{\pi } }}{\left( {\frac{1}{{{\mathrm{a}_0}}}} \right)^{3/2}}{\mathrm{e}^{ - \mathrm{r}/{\mathrm{a}_0}}}, , where a0{a_0} is Bohr's radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is:
View answer and explanation
Correct option: C
ψ24πr2dr{\mathrm{\psi }^2}4\mathrm{\pi }{r^2}dr
Option C (ψ24πr2dr{\mathrm{\psi }^2}4\mathrm{\pi }{r^2}dr) is correct because it represents the probability of finding the electron within a spherical shell of radius r and thickness dr.
475 Hard
475 of 492
Application +4 | -1
The velocities of two particles A and B are 0.05 and 0.02   ms−1\;\mathrm{m}{\mathrm{s}^{ - 1}} respectively. The mass of B is five times the mass of A. The ratio of their de-Broglie’s wavelength is
View answer and explanation
Correct option: A
2:12:1
Option A (2:12:1) is correct because the ratio of de Broglie wavelengths is inversely proportional to the product of mass and velocity, and considering the given mass and velocity ratios.
476 Easy
476 of 492
Analyzing +4 | -1
The atomic number of the element having maximum number of unpaired 3p-electrons is:
View answer and explanation
Correct option: A
15
Option A (15) is correct because the element with atomic number 15 (Phosphorus) has the electronic configuration [Ne] 3s² 3p³, resulting in three unpaired 3p electrons, which is the maximum possible.
477 Easy
477 of 492
Application +4 | -1
The electron in the hydrogen atom is excited to the 4th orbit. Number of spectral lines expected in the emission is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the number of spectral lines is calculated using the formula n(n-1)/2 where n is the orbit number.
478 Easy
478 of 492
Understanding +4 | -1
Which relates to light only as stream of particles?
View answer and explanation
Correct option: B
Photoelectronic effect
Option B (Photoelectronic effect) is correct because the photoelectric effect demonstrates the particle nature of light (photons) interacting with electrons.
479 Hard
479 of 492
Application +4 | -1
What should be the mass of the sodium photon if its wavelength is 5894 A∘ ,5894 \, \overset{\circ }{\mathop{\text{A}}}\,, the velocity of light is 3×1083 \times 10^8 metre/second and the value of h\text{h} is 6.652×10−34 kg m2/sec?6.652 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{sec}?
View answer and explanation
Correct option: D
3.746×10−363.746 \times 10^{-36}
Option D (3.746×10−363.746 \times 10^{-36}) is correct because it is the calculated mass of the sodium photon using the de Broglie wavelength equation and the relationship between energy, mass, and the speed of light.
480 Hard
480 of 492
Application +4 | -1
If the shortest wavelength of H-atom in Lyman series is x, the longest wavelength in Balmer series of He+\mathrm{H}{\mathrm{e}^ + } is
View answer and explanation
Correct option: D
9x5\frac{{9x}}{5}
Option D (9x5\frac{{9x}}{5}) is correct because the shortest wavelength in the Lyman series corresponds to the transition from n=∞ to n=1, and the longest wavelength in the Balmer series of He+He^+ corresponds to the transition from n=3 to n=2. By equating the Rydberg formula for both, we can find the relationship between x and the required wavelength.
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