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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Showing 241-270 of 492 questions

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241 Easy
241 of 492
Remembering +4 | -1
A node is a surface on which the probability of finding an electron is:
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because a node is defined as a region in space where the probability of finding an electron is zero.
242 Easy
242 of 492
Understanding +4 | -1
For which species, Bohr's theory does not apply:
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Correct option: B
Be
Option B (Be) is correct because Bohr's theory is only applicable to single-electron species, and Be has 4 electrons.
243 Hard
243 of 492
Understanding +4 | -1
The mass defect of the nuclear reaction: 5B8→4Be8+e+{ }_5 {B}^8 \rightarrow{ }_4 {Be}^8+{e}^{+}is
View answer and explanation
Correct option: D
m={m}= At. mass of 5B8−{ }_5 {B}^8- At. mass of 4Be8+{ }_4 {Be}^8+ mass of two electron
Option D (m={m}= At. mass of 5B8−{ }_5 {B}^8- At. mass of 4Be8+{ }_4 {Be}^8+ mass of two electron) is correct because the mass defect includes the mass of the positron emitted, which accounts for the difference in nuclear charge and requires adding two electron masses to balance the equation.
244 Medium
244 of 492
Understanding +4 | -1
The formation of energy bonds in solids are in accordance with
View answer and explanation
Correct option: B
Bohr’s theory
Option B (Bohr’s theory) is correct because Bohr's theory explains the discrete energy levels and transitions of electrons, which leads to the formation of energy bands in solids.
245 Medium
245 of 492
Understanding +4 | -1
Electron density in the YZ plane of 3dx2−y23{d_{{x^2} - {y^2}}} orbital is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the 3dx2−y23d_{x^2-y^2} orbital lies along the x and y axes; therefore, the electron density in the YZ plane, which is perpendicular to the x-axis, is zero.
246 Hard
246 of 492
Understanding +4 | -1
Which has highest e/m ratio?
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Correct option: B
H+{\mathrm{H}^ + }
Option B (H+{\mathrm{H}^ + }) is correct because it has the highest charge-to-mass ratio due to having a charge of +1 and the smallest mass among the given options.
247 Medium
247 of 492
Understanding +4 | -1
A certain orbital has no angular nodes and two radial nodes. The orbital is
View answer and explanation
Correct option: B
3s3s
Option B (3s3s) is correct because a 3s orbital has two radial nodes and zero angular nodes.
248 Hard
248 of 492
Understanding +4 | -1
The path of deflection of electron beam is:
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Correct option: B
Inversely proportional to the magnitude of applied magnetic field
Option A (Directly proportional to the magnitude of applied magnetic field) is correct because the force on a moving charge in a magnetic field is directly proportional to the field strength, leading to greater deflection.
249 Hard
249 of 492
Application +4 | -1
The wave number (V‾)\left( {\overline V } \right) in terms of Rydberg’s constant for an electronic transition between two levels of Li2+L{i^{2 + }} ion, whose sum is four and difference is 2 will be
View answer and explanation
Correct option: A
8R
Option A (8R) is correct because the wave number is proportional to Z2Z^2 and the energy difference is determined by the principal quantum numbers.
250 Easy
250 of 492
Remembering +4 | -1
The photoelectric effect occurs only when the incident light has more frequency than a certain minimum:
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Correct option: A
Frequency
Option A (Frequency) is correct because the photoelectric effect requires the incident light to have a frequency above the threshold frequency for the given material.
251 Medium
251 of 492
Remembering +4 | -1
Which of the following conditions is incorrect for a well behaved wave function (ψ)(\psi )
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Correct option: C
ψ\psi must be infinite
Option C (ψ\psi must be infinite) is correct because a well-behaved wave function must be finite, not infinite.
252 Easy
252 of 492
Application +4 | -1
Which of the following set of quantum numbers represents the highest energy orbital in a multi electron atom ?
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Correct option: D
n=4,l=1,m=0,s=−12n = 4,l = 1,m = 0,s = - \frac{1}{2}
Option D (n=4,l=1,m=0,s=−12n = 4,l = 1,m = 0,s = - \frac{1}{2}) is correct because the energy of an orbital in a multi-electron atom primarily depends on the sum of n+l, and when n+l is the same, the orbital with the higher n value has higher energy.
253 Medium
253 of 492
Application +4 | -1
If the radius of first Bohr's orbit be a0{a_0} , then the radius of third Bohr's orbit would be:
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Correct option: C
9×a09 \times {a_0}
Option C (9×a09 \times {a_0}) is correct because the radius of the nth Bohr orbit is proportional to n2n^2.
254 Hard
254 of 492
Application +4 | -1
The ionisation enthalpy of hydrogen atom is 1.312×106Jmol−11.312 \times {10^6}\mathrm{Jmo}{\mathrm{l}^{ - 1}}. The energy required to excite the electron in the atom from n1=1  to  n2=2{n_1} = 1\;\mathrm{to }\;{n_2} = 2 is
View answer and explanation
Correct option: D
9.84×105  J  mol−19.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}}
Option D (9.84×105  J  mol−19.84 \times {10^5}{\;J\;mo}{\mathrm{l}^{ - 1}}) is correct because it represents the energy difference between the n=1 and n=2 energy levels in a hydrogen atom, calculated using the ionization enthalpy.
255 Medium
255 of 492
Understanding +4 | -1
Among the following, the correct statement about cathode ray discharge tube is
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Correct option: B
in the absence of external electrical or magnetic field, cathode rays travel in straight lines.
Option B is correct because in the absence of external fields, cathode rays, being streams of electrons, travel in straight lines.
256 Easy
256 of 492
Application +4 | -1
In a Hydrogen atom, if the energy of electron in the ground state is −x  ev,- x\,\,ev, then that in the 2nd excited state of He+H{e^ + } is
View answer and explanation
Correct option: B
−49x  ev- \frac{4}{9}x\,\,ev
Option B (−49x, ev- \frac{4}{9}x\\,\,ev) is correct because the energy is inversely proportional to n squared and directly proportional to Z squared.
257 Easy
257 of 492
Application +4 | -1
The value of Planck’s constant is 6.63×10−34  Js.6.63 \times {10^{ - 34}}\;\mathrm{Js}.. The velocity of light is 3.0×108  m  s−1.3.0 \times {10^8}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}.. Which value is closest to the wavelength in nanometre of a quantum of light with frequency of 8×1015  s−1?8 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}?
View answer and explanation
Correct option: C
4×10−84 \times {10^{ - 8}}
Option C is correct. Using the formula λ = c/f, the calculated wavelength is approximately 4 × 10-8 meters, which is 40 nm.
258 Easy
258 of 492
Understanding +4 | -1
Which consists of particle of matter?
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Correct option: D
All of these
Option D (All of these) is correct because alpha rays (helium nuclei), beta rays (electrons), and cathode rays (electrons) all consist of particles of matter.
259 Medium
259 of 492
Remembering +4 | -1
If ‘r’ is the radius of first orbit, the radius of nth orbit of H-atom is given by
View answer and explanation
Correct option: A
rn2r{n^2}
Option A (rn2r{n^2}) is correct because the radius of the nth orbit in a hydrogen atom is proportional to n2n^2, where n is the principal quantum number.
260 Easy
260 of 492
Understanding +4 | -1
Which one of the following set of ions represents a collection of isoelectronic species
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Correct option: B
CN−, C2−2, N2C{N^ - },\,C_2^{ - 2},\,{N_2}
Option B (CN−, C2−2, N2C{N^ - },\,C_2^{ - 2},\,{N_2}) is correct because all three species have 14 electrons, making them isoelectronic.
261 Medium
261 of 492
Application +4 | -1
The Bohr’s orbit radius for the hydrogen atom (n=1) is approximately 0.53 Å. The radius for the first excited state (n=2) orbit is
View answer and explanation
Correct option: C
2.12 Å
Option C (2.12 Å) is correct because the radius of the nth orbit is proportional to n2n^2, so the radius of the first excited state (n=2) is 4 times the radius of the ground state (n=1).
262 Easy
262 of 492
Application +4 | -1
The ratio of energy of a photon of 2000A02000{A^0} wavelength 396 to that of 4000A04000{A^0} 396 is
View answer and explanation
Correct option: D
2
Option D (2) is correct because the energy of a photon is inversely proportional to its wavelength. Therefore, halving the wavelength doubles the energy.
263 Medium
263 of 492
Analyzing +4 | -1
Assertion: In absorption spectrum, nf>nin_f > n_i, while in emission spectrum, nf>nin_f > n_i

Reason: Energy is limited in absorption spectrum, that is, electron moves from lower to higher orbit and energy is absorbed in emission spectrum, because election jumps from higher to lower orbit.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because the assertion incorrectly states the relationship between initial and final energy levels in absorption and emission spectra, and the reason provides an inaccurate explanation of these processes.
264 Easy
264 of 492
Analyzing +4 | -1
Which of the following pairs are isoelectronic and isotopic.
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Correct option: A
39K+   and  40K+^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }
Option A (39K+   and  40K+^{39}{K^ + }\,\,\,\mathrm{and}\,{\,^{40}}{K^ + }) is correct because they have the same number of electrons (isoelectronic) but different mass numbers (isotopic).
265 Easy
265 of 492
Application +4 | -1
The number of neutrons in the dipositive zinc ion (A of Zn = 65)
View answer and explanation
Correct option: A
35
Option A (35) is correct because the number of neutrons is calculated by subtracting the atomic number (number of protons) from the mass number.
266 Hard
266 of 492
Application +4 | -1
If pp is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ,\lambda , then for 1.5 p momentum of the photoelectron, the wavelength of the light should be
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)
View answer and explanation
Correct option: A
49λ\frac{4}{9}\lambda
Option A (49λ\frac{4}{9}\lambda) is correct because the kinetic energy is proportional to p2p^2, and the wavelength is inversely proportional to the square root of the kinetic energy.
267 Hard
267 of 492
Application +4 | -1
The uncertainity in position for a dust particle m=10−11g;   diameter=10−4m = {10^{ - 11}}g;\,\,\,diameter = {10^{ - 4}} cm and velocity =10−4 cm/s= {10^{ - 4}}\,cm/s) will be (The error in measurement of velocity is 1%)
View answer and explanation
Correct option: C
5.27×10−65.27 \times {10^{ - 6}} cm
Option C (5.27×10−65.27 \times {10^{ - 6}} cm) is correct because it is the calculated uncertainty in position using the Heisenberg uncertainty principle.
268 Easy
268 of 492
Understanding +4 | -1
Which set of phenomenon shown by the radiation proves the dual nature of radiation?
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Correct option: C
Interference and photoelectric effect
Option C (Interference and photoelectric effect) is correct because interference demonstrates the wave nature and the photoelectric effect demonstrates the particle nature of radiation.
269 Medium
269 of 492
Remembering +4 | -1
The radiation having maximum wavelength is
View answer and explanation
Correct option: B
Radio waves
Option B (Radio waves) is correct because radio waves have the longest wavelength in the electromagnetic spectrum.
270 Hard
270 of 492
Analyzing +4 | -1
Statement (S): Number of sub-shells in a shell is equal to the number of shell.

Explanation (E): According to Sommerfeld: nk= Length of major axis  Length of minor axis \frac{n}{k}=\frac{\text { Length of major axis }}{\text { Length of minor axis }}
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option A is correct because the statement that the number of sub-shells in a shell equals the shell number is correct, and Sommerfeld's equation explains the elliptical orbits which lead to subshells.
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