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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 592 questions

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271 Easy
271 of 592
Understanding +4 | -1
Which of the following energy level may bring absorption of photon but never emission of photon?
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Correct option: C
1s1 s
Option C (1s1s) is correct because the 1s energy level is the lowest and thus cannot emit a photon, only absorb.
272 Medium
272 of 592
Application +4 | -1
What is the ratio of maximum λ\lambda of Lyman and Balmer series
View answer and explanation
Correct option: A
46508
Option A (May-27) is correct and should be 27/5, because it is calculated by finding the ratio of the maximum wavelengths of the Lyman and Balmer series using the Rydberg formula.
273 Easy
273 of 592
Understanding +4 | -1
The quantum number not obtained from the schrodinger wave equation is
View answer and explanation
Correct option: D
s
The correct answer is D (s) because the spin quantum number (s) is not derived directly from solving the Schrodinger wave equation.
274 Easy
274 of 592
Remembering +4 | -1
The shape of an orbital is given by the quantum number
View answer and explanation
Correct option: B
ll
Option B (ll) is correct because the shape of an orbital is determined by the azimuthal quantum number (ll).
275 Easy
275 of 592
Remembering +4 | -1
The velocity of a photon is:
View answer and explanation
Correct option: A
Independent of its wavelength
Option A (Independent of its wavelength) is correct because the velocity of a photon in a vacuum is a constant, the speed of light (c), and is independent of its wavelength or frequency.
276 Easy
276 of 592
Understanding +4 | -1
The atom with lowest atomic number that has a ground state (EC) of (n−1)d6ns2(n-1) d^6 n s^2 is in the
View answer and explanation
Correct option: C
4th period
Option C (4th period) is correct because the electronic configuration (n−1)d6ns2(n-1) d^6 n s^2 indicates a transition metal, and the d-orbitals are first filled in the 4th period (n=4).
277 Easy
277 of 592
Application +4 | -1
The de Broglie's wavelength of electron present in first Bohr orbit of H\mathrm{H} atom is
View answer and explanation
Correct option: B
2π×0.529 A∘2 \pi \times 0.529 \, \mathop {\text{A}}\limits^ \circ
The correct answer is B; 2π×0.529 A∘2 \pi \times 0.529 \, \mathop{{\text{A}}}\limits^ \circ. The de Broglie wavelength is equal to the circumference of the first Bohr orbit.
278 Medium
278 of 592
Analyzing +4 | -1
In Bohr's model of the hydrogen atom, let r,vr, v and EE represent the orbit radius, speed of an electron and the total energy of the electron, respectively. Which of following relation is proportional to the orbit number nn ?
View answer and explanation
Correct option: A
v⋅rv \cdot r
The correct answer is C (KI) because it forms a yellow-brown precipitate with copper sulfate that turns white with excess sodium thiosulfate, and a yellow precipitate with silver ions insoluble in ammonium hydroxide.
279 Easy
279 of 592
Application +4 | -1
According to de-Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g100 \mathrm{~g} does not move like a wave when it is thrown by a bowler at a speed of 100 km/h100 \mathrm{~km} / \mathrm{h}. Calculate the wavelength of the ball.
View answer and explanation
Correct option: C
238.5×10−36 m238.5 \times 10^{-36} \mathrm{~m}
Option C (238.5×10−36 m238.5 \times 10^{-36} \mathrm{~m}) is correct because it is calculated using the de Broglie wavelength formula with the given mass and velocity.
280 Easy
280 of 592
Remembering +4 | -1
The number of protons, neutrons and electrons in 71175Lu{}_{71}^{175}\mathrm{Lu} , respectively, are
View answer and explanation
Correct option: A
71,  104  and  71\mathrm{71,}\,\,\mathrm{104}\,\,\mathrm{and}\,\,\mathrm{71}
Option A (71, 104, and 71) is correct because the number of protons and electrons equals the atomic number, and the number of neutrons is the mass number minus the atomic number.
281 Easy
281 of 592
Analyzing +4 | -1
Which statement relating to the spectrum of H atom is false?
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Correct option: C
The spectral lines are closer together at longer wavelengths
Option C (The spectral lines are closer together at longer wavelengths) is false because the spectral lines in the hydrogen atom spectrum converge at shorter wavelengths (higher energies), not longer wavelengths.
282 Easy
282 of 592
Understanding +4 | -1
The angular momentum of an electron in a given stationary state can be expressed as mevr=nh2πm_e v r=n \frac{h}{2 \pi}. Based on this expression an electron can move only in those orbits for which its angular momentum is
View answer and explanation
Correct option: B
Integral multiple of h2π\frac{h}{2 \pi}
Option B (Integral multiple of h2π\frac{h}{2 \pi}) is correct because Bohr's quantization condition states that the angular momentum of an electron is an integral multiple of h2π\frac{h}{2 \pi}.
283 Medium
283 of 592
Remembering +4 | -1
The ratio of the radii of the first three Bohr orbits is :
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Correct option: C
1:4:91: 4: 9
Option C (1: 4: 9) is correct because the radius of a Bohr orbit is proportional to n2n^2, where n is the principal quantum number.
284 Medium
284 of 592
Application +4 | -1
The quantum number of four electrons are given below:
I. n=4,  l=2,  ml=−2,  ms=−12n = 4,\,\,l = 2,\,\,{m_l} = - 2,\,\,{m_s} = - \frac{1}{2}
II. n=3,  l=2,  ml=1,  ms=+12n = 3,\,\,l = 2,\,\,{m_l} = 1,\,\,{m_s} = + \frac{1}{2}
III. n=4,  l=1,  ml=0,  ms=+12n = 4,\,\,l = 1,\,\,{m_l} = 0,\,\,{m_s} = + \frac{1}{2}
IV. n=3,  l=1,  ml=1,  ms=−12n = 3,\,\,l = 1,\,\,{m_l} = 1,\,\,{m_s} = - \frac{1}{2}
The correct order of their increasing energies will be
View answer and explanation
Correct option: C
IV<II<III<I \mathrm{IV < II < III < I}\,
The correct order is IV < II < III < I because the energy of an electron is determined by the (n+l) rule, and if (n+l) is the same, the electron with lower n has lower energy.
285 Medium
285 of 592
Application +4 | -1
A dye absorb light of λ=4530Ao\lambda=4530 \mathop {\text{A}}\limits^{\text{o}} and then fluorescence of 5080Ao5080 \mathop {\text{A}}\limits^{\text{o}} Assuming that under given conditions 47%C47 \% \mathrm{C} absorbed energy is re-emitted out as fluorescence, calculate ratio of quanta emitted out to the number of quanta absorb
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Correct option: A
0.5270.527
Option A (0.527) is correct because it represents the ratio of emitted to absorbed quanta, calculated using the wavelengths and the given energy conversion efficiency.
286 Medium
286 of 592
Application +4 | -1
If nucleus and atom are considered as perfect spheres with diameters 4×10−15 m4 \times 10^{-15} \mathrm{~m} and 2×10−102 \times 10^{-10} m\mathrm{m}, respectively, then the ratio of the volumes of nucleus and atom should be
View answer and explanation
Correct option: B
8×10−15:18 \times 10^{-15}: 1
Option B (8×10−15:18 \times 10^{-15}: 1) is correct because the ratio of volumes is proportional to the cube of the ratio of their diameters.
287 Hard
287 of 592
Application +4 | -1
In a relative scale the deBrogle wave length of moving neutron at 27∘C27^{\circ} \mathrm{C} is y1; A0\mathrm{y}_1 ; \mathrm{~A}^0 and the de-Broglees wave length of moving SO2\mathrm{SO}_2 molecules at 927∘C927^{\circ} \mathrm{C} is y2 A0\mathrm{y}_2 \mathrm{~A}^0 then (y1y2)=\left(\frac{\mathrm{y}_1}{\mathrm{y}_2}\right)=? (Assume that mass of neutron &\& proton is same) (mol. Wt of SO2=64\mathrm{SO}_2=64)
View answer and explanation
Correct option: A
16
Option A (16) is correct because it is the ratio of the de Broglie wavelengths calculated using the given temperatures and masses.
288 Medium
288 of 592
Understanding +4 | -1
Statement I : The amount of light emitted (intensity of radiation) from a black body and its spectral distribution depends only on its temperature
Statement II : When a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength
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Correct option: A
Both Statement I and Statement II are correct
Option A is correct because both statements are fundamental principles in physics: black body radiation depends on temperature, and shorter wavelengths bend more in a prism.
289 Hard
289 of 592
Application +4 | -1
The ratio of angular momentum of electron in two successive orbits is a (a>1)a \, (a>1) and their difference is b,b, then ab\frac{a}{b} is equal to
View answer and explanation
Correct option: D
(n+1n)⋅2πh\left(\frac{n+1}{n}\right) \cdot \frac{2 \pi}{h}
The correct answer is D ((n+1n)⋅2πh\left(\frac{n+1}{n}\right) \cdot \frac{2 \pi}{h}) because it correctly represents the ratio a/b based on the angular momentum in successive orbits.
290 Easy
290 of 592
Understanding +4 | -1
Which electron transition in a hydrogen atom requires the largest amount of energy?
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Correct option: A
From n=1 to n=2
Option A (From n=1 to n=2) is correct because the energy difference between n=1 and n=2 is the largest in the hydrogen atom spectrum.
291 Medium
291 of 592
Remembering +4 | -1
The total energy in 1st 1^{\text {st }} orbit of hydrogen atom is given by
View answer and explanation
Correct option: D
−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}
Option D (−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}) is correct because it represents the total energy of an electron in the first orbit of a hydrogen atom, derived from Bohr's model.
292 Easy
292 of 592
Understanding +4 | -1
Which of the following expression represent the wave number of spectral lines in Balmer series (if nn is the principal quantum number of higher energy level)?
View answer and explanation
Correct option: C
R(n−2)(n+2)4n2\frac{R(n-2)(n+2)}{4 n^2}
The correct answer is C. The wave number for the Balmer series is given by R(n−2)(n+2)4n2\frac{R(n-2)(n+2)}{4 n^2}, derived from the Rydberg formula.
293 Hard
293 of 592
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 it is derived from using the Rydberg formula for the shortest wavelength in the Lyman series of H and the longest wavelength in the Balmer series of He+.
294 Medium
294 of 592
Remembering +4 | -1
In cathode ray discharge tube experiment, when sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles from negative electrode to positive electrodes. These particles are called
View answer and explanation
Correct option: D
Both (2) or (3)
Option D (Both (2) or (3)) is correct because the particles emitted from the negative electrode in a cathode ray tube are called cathode rays or cathode ray particles.
295 Easy
295 of 592
Application +4 | -1
Which of the following pairs of ions have the same electronic configuration?
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Correct option: B
Fe3+,Mn2+\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}
Option B (Fe3+,Mn2+\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}) is correct because both ions have the same electronic configuration of [Ar]3d⁵.
296 Easy
296 of 592
Application +4 | -1
Hydrogen atoms are excited to n=4n = 4 state, In the spectrum of emitted radiation, number of lines in the
UV and visible regions are resp.
View answer and explanation
Correct option: D
3,23,2
Option D (3,23,2) is correct because when hydrogen atoms are excited to n=4, there are 3 Lyman series lines (UV region) and 2 Balmer series lines (visible region).
297 Medium
297 of 592
Application +4 | -1
What is the de-Broglie wavelength (in nm{nm} ) of an electron in the ground state of hydrogen atom is ?

 [K.E. =13.6 eV;1 eV=1.602×10−19 J ] \text { [K.E. }=13.6 \,{eV} ; 1 \,{eV}=1.602 \times 10^{-19} {~J} \text { ] }
View answer and explanation
Correct option: A
0.33
Option A (0.33) is correct because it is the de-Broglie wavelength calculated using the kinetic energy of the electron.
298 Medium
298 of 592
Understanding +4 | -1
Which of the following does not characterise X-rays?
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Correct option: C
Deflected by electric and magnetic fields
Option C (Deflected by electric and magnetic fields) is correct because X-rays are electromagnetic radiation and are not deflected by electric or magnetic fields.
299 Medium
299 of 592
Understanding +4 | -1
The ratio of the radius of the orbit for the electron orbiting the hydrogen nucleus to that of an electron orbiting a deuterium nucleus is:
View answer and explanation
Correct option: A
1∶1
The correct answer is A (1∶1) because the radius of the orbit depends on the atomic number (Z) and the principal quantum number (n), and since both hydrogen and deuterium have Z=1, the ratio is 1:1.
300 Easy
300 of 592
Remembering +4 | -1
According to Heisenberg uncertainty principle:
View answer and explanation
Correct option: B
Δx×Δp≥h4π\Delta x \times \Delta p \geq \frac{h}{4 \pi}
Option B (Δx×Δp≥h4π\Delta x \times \Delta p \geq \frac{h}{4 \pi}) is correct because the Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum is greater than or equal to h4π\frac{h}{4 \pi}.
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