Back to STRUCTURE OF ATOM
Your Progress
0 / 592 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • STRUCTURE OF ATOM
Structure of atom practice · Mixed difficulty
Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.
Practice format and chapter quiz
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.
Take a chapter quiz →
Showing 541-570 of 592 questions
541
Medium
541 of 592
Remembering
+4 | -1
Which of the following statement is/are correct about Hertz ?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given statements about Hertz (Hz) are accurate and well-established.
542
Easy
542 of 592
Analyzing
+4 | -1
Statement (S): Specific charge of -particles is twice to that of proton.
Explanation (E): Specific charge is given by
Explanation (E): Specific charge is given by
View answer and explanation
Correct option: B
S is wrong but E is correct
Option B (S is wrong but E is correct) is correct because the specific charge of an alpha particle is half that of a proton, and the explanation correctly defines specific charge.
543
Medium
543 of 592
Remembering
+4 | -1
Cathode rays have same charge to mass ratio as
View answer and explanation
Correct option: A
-rays
Option A (-rays) is correct because cathode rays are streams of electrons, and beta particles are also electrons emitted from the nucleus.
544
Medium
544 of 592
Application
+4 | -1
The ratio of wavelength values of series limit lines of Balmer series and Paschen series are
View answer and explanation
Correct option: A
0.172916667
Option A (04:09) is correct because the ratio of the wavelengths is found by using the Rydberg formula for the series limit of the Balmer and Paschen series.
545
Medium
545 of 592
Application
+4 | -1
The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The possible energy value of electron in the excited state of is
View answer and explanation
Correct option: C
Option C (-30.6 eV) is correct because it's calculated using the formula for energy levels in hydrogen-like atoms, considering the excited state (n=2) and the atomic number of Li2+ (Z=3).
546
Medium
546 of 592
Application
+4 | -1
Electromagnetic radiations of wavelength 240 are just sufficient to ionize sodium atom. The ionization energy of sodium (in ) is
View answer and explanation
Correct option: B
498.58
Option B (498.58) is correct because the ionization energy is calculated using the given wavelength and the formula I.E. = hc/λ.
547
Easy
547 of 592
Application
+4 | -1
Which of the following particles has more electrons than neutrons
View answer and explanation
Correct option: C
Option C () is correct because it has 10 electrons and 8 neutrons, thus having more electrons than neutrons.
548
Easy
548 of 592
Understanding
+4 | -1
The H-spectrum show
View answer and explanation
Correct option: D
Presence of quantised energy level
Option D (Presence of quantised energy level) is correct because the hydrogen spectrum provides evidence for the existence of discrete, quantized energy levels within the hydrogen atom.
549
Hard
549 of 592
Application
+4 | -1
What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to
View answer and explanation
Correct option: B
Option B () is correct because the de Broglie wavelength is calculated using the formula , where m is the mass of one proton and v is the velocity.
550
Easy
550 of 592
Remembering
+4 | -1
In case of orbital:
View answer and explanation
Correct option: C
Probability of finding the electron is maximum along
x and y-axes
x and y-axes
Option C (Probability of finding the electron is maximum along x and y-axes) is correct because the orbital has lobes along the x and y axes, indicating a higher probability of finding electrons along those axes.
551
Easy
551 of 592
Remembering
+4 | -1
The electronic energy levels of the hydrogen atom in the Bohr^' s theory are called:
View answer and explanation
Correct option: B
Orbits
Option B (Orbits) is correct because Bohr's theory postulates that electrons revolve around the nucleus in specific, quantized orbits.
552
Medium
552 of 592
Application
+4 | -1
The principal quantum number of H-atom orbital, if the electron energy is -3.4 eV, will be
View answer and explanation
Correct option: B
2
Option B (2) is correct because the energy of an electron in a hydrogen atom is given by -13.6/n² eV, and solving for n when the energy is -3.4 eV gives n=2.
553
Easy
553 of 592
Application
+4 | -1
If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by
View answer and explanation
Correct option: A
Option A () is correct because it is derived from the Rydberg formula for the wavelength of emitted radiation when an electron transitions between energy levels.
554
Medium
554 of 592
Application
+4 | -1
A typical golf ball weighs . If it is moving with a velocity of , its de-Broglie wavelength is
View answer and explanation
Correct option: C
Option B () is correct because it is calculated using the de Broglie wavelength formula, .
555
Easy
555 of 592
Understanding
+4 | -1
Which is not an atomic orbital?
View answer and explanation
Correct option: A
2d
Option A (2d) is correct because the 2d orbital does not exist according to quantum mechanical principles.
Explore Other Chapters in CHEMISTRY
View All →
556
Easy
556 of 592
Application
+4 | -1
The energy in joule corresponding to light of wavelength 45 nm is
View answer and explanation
Correct option: D
Option D () is correct because the energy is calculated using the Planck's constant, speed of light, and the wavelength of the light.
557
Medium
557 of 592
Application
+4 | -1
Bond dissociation on energy of is . The longest wavelength of photon that can break this bond would be
View answer and explanation
Correct option: C
Option C () is correct because it is the longest wavelength of a photon that can break the bond, calculated using the energy-wavelength relationship.
558
Medium
558 of 592
Application
+4 | -1
According to Heisenberg's uncertainty principle, the product of uncertainties in position and velocities for an electron of mass is
View answer and explanation
Correct option: C
Option C () is correct because it is calculated using Heisenberg's uncertainty principle.
559
Easy
559 of 592
Application
+4 | -1
A metallic ion ion has a electronic configuration of 2, 8, 14 and the ionic weight is 56 amu. The number of neutrons in its nucleus are
View answer and explanation
Correct option: A
30
Option A (30) is correct because the number of neutrons is the difference between the mass number and the atomic number.
560
Hard
560 of 592
Application
+4 | -1
Energy required to stop the ejection of electrons from plate is . If radiation of strikes the plate, the work function is
View answer and explanation
Correct option: A
Option A () is correct because the work function is calculated by subtracting the stopping potential energy from the photon energy.
561
Medium
561 of 592
Application
+4 | -1
Quantum numbers for some electrons are given below
A : B : C: D:
The correct increasing order of energy of electrons
A : B : C: D:
The correct increasing order of energy of electrons
View answer and explanation
Correct option: C
D B C A
Option C (D B C A) is correct because it represents the increasing order of energy based on the (n+l) rule and the n value when (n+l) is the same.
562
Medium
562 of 592
Application
+4 | -1
The radius of an orbit of H – atom is . Then the angular momentum of an electron in this orbit is
View answer and explanation
Correct option: B
Option B () is correct because the radius is used to determine the orbit number (n=3), and then the angular momentum is calculated using the formula .
563
Easy
563 of 592
Remembering
+4 | -1
Which among the following radiations has maximum frequency?
View answer and explanation
Correct option: C
-rays
Option C (-rays) is correct because gamma rays have the highest frequency among the given options in the electromagnetic spectrum.
564
Easy
564 of 592
Remembering
+4 | -1
How many neutrons are present in tritium nucleus?
View answer and explanation
Correct option: A
2
Option A (2) is correct because tritium, an isotope of hydrogen, has 1 proton and 2 neutrons in its nucleus (mass number 3, atomic number 1).
565
Medium
565 of 592
Application
+4 | -1
A photon of light of wavelength 6000 Å has energy . What will be the wavelength of photon of a light which has energy of photon ?
View answer and explanation
Correct option: A
1500 Å
Option A (1500 Å) is correct because energy and wavelength are inversely proportional, so quadrupling the energy reduces the wavelength to one-fourth.
566
Medium
566 of 592
Understanding
+4 | -1
ASSERTION (A) : Two electrons in an atom can have the same values of four quantum numbers.
REASON (R) : Two electrons in an atom can be present in the same shell, subshell and orbital and have the same spin.
REASON (R) : Two electrons in an atom can be present in the same shell, subshell and orbital and have the same spin.
View answer and explanation
Correct option: D
If and both are wrong.
Option D (If A and R both are wrong) is correct because Assertion (A) is wrong due to Pauli Exclusion Principle, and Reason (R) is also wrong because electrons in the same orbital must have opposite spins.
567
Medium
567 of 592
Remembering
+4 | -1
The principal and azimuthal quantum number of electrons in orbitals are
View answer and explanation
Correct option: C
4,3
Option C (4,3) is correct because for a 4f orbital, the principal quantum number (n) is 4 and the azimuthal quantum number (l) is 3.
568
Medium
568 of 592
Application
+4 | -1
Based on the equation
the wavelength of the light that must be absorbed to excite hydrogen electron from level to level will be
the wavelength of the light that must be absorbed to excite hydrogen electron from level to level will be
View answer and explanation
Correct option: A
Option A () is correct because it is the calculated wavelength of light absorbed during the electron transition from n=1 to n=2 in a hydrogen atom, using the provided energy equation.
569
Easy
569 of 592
Understanding
+4 | -1
Which one of the following set of ions represents a collection of isoelectronic species
View answer and explanation
Correct option: B
Option B () is correct because all three species have the same number of electrons (14).
570
Medium
570 of 592
Analyzing
+4 | -1
Statement-I: Emitted radiation will be joule when an electron jump from to in -atom.
Statement-II: Balmer series radiations belong to IR region for hydrogen atom only.
Statement-II: Balmer series radiations belong to IR region for hydrogen atom only.
View answer and explanation
Correct option: B
Statement I and Statement II both are incorrect
The correct answer is B (Statement I and Statement II both are incorrect) because the emitted radiation energy is 4.08 × 10-19 J, and Balmer series radiations belong to the visible region.
Revision notes and key takeaways
Review STRUCTURE OF ATOM in your textbook or class notes, then use this set to check your understanding.
Showing 541-570 of 592 questions
Continue Your Practice
📚 Switch Set (ALL)
0
/592
0✓
0✗
0% Accuracy
