The three quantum numbers n,l $n,l$ and $m$ are the outcome of:

Heisenberg’s uncertainty principle

STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 2

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Application +4 | -1

Which of the following particles moving with same velocity would be associated with smaller de-Broglie wavelength?

View answer and explanation

Correct option: B

Oxygen molecule

Option B (Oxygen molecule) is correct because the de Broglie wavelength is inversely proportional to the mass, and oxygen molecule has the largest mass among the given options.

Solution approach

Correct Answer + Core Reason: Oxygen molecule. The de-Broglie wavelength is inversely proportional to the mass of the particle.

Step-by-Step Solution:

Step 1: Recall the de-Broglie wavelength formula: $\lambda = \frac{h}{mv}$ , where h is Planck's constant, m is the mass of the particle, and v is its velocity.

Step 2: Since the velocity is the same for all particles, the de-Broglie wavelength is inversely proportional to the mass: $\lambda \propto \frac{1}{m}$ .

Step 3: Compare the masses of the given particles:
- Helium molecule ( $He_2$ ): approximately 8 amu
- Oxygen molecule ( $O_2$ ): approximately 32 amu
- Hydrogen molecule ( $H_2$ ): approximately 2 amu
- Carbon molecule ( $C_x$ ): This option is ambiguous. Assuming it is a $C_2$ molecule, it would be approximately 24 amu.

Step 4: Since the oxygen molecule has the largest mass, it will have the smallest de-Broglie wavelength.

Why other options are wrong:
- Option A (Helium molecule): Helium molecule has a smaller mass than oxygen molecule, so it will have a larger de-Broglie wavelength.
- Option C (Hydrogen molecule): Hydrogen molecule has the smallest mass, so it will have the largest de-Broglie wavelength.
- Option D (Carbon molecule): Assuming it is $C_2$ , it has smaller mass than oxygen molecule, so it will have a larger de-Broglie wavelength.

Key Concepts:
- de-Broglie wavelength
- Mass of molecules

Memory Aids:
- de-Broglie wavelength: $\lambda = \frac{h}{mv}$

Quick Tips:
- Remember that de-Broglie wavelength is inversely proportional to mass when velocity is constant.
- Know the approximate masses of common molecules.

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Which one of the following has unit positive charge and 1 u mass?

View answer and explanation

Correct option: C

Proton

Option C (Proton) is correct because a proton has a unit positive charge and a mass of approximately 1 atomic mass unit (1 u).

Solution approach

Correct Answer + Core Reason: Proton. A proton has a +1 charge and a mass of approximately 1 u.

Step-by-Step Solution:

Step 1: Recall the properties of subatomic particles.

Step 2: Consider the charge and mass of each particle:
- Electron: Negative charge, negligible mass.
- Neutron: No charge, mass of approximately 1 u.
- Proton: Positive charge, mass of approximately 1 u.

Step 3: Identify the particle with unit positive charge and 1 u mass.

Step 4: The proton fits the description.

Why other options are wrong:
- Option A: Electron has a negative charge, not positive.
- Option B: Neutron has no charge.
- Option D: Since proton is the correct answer, none of these is incorrect.

Key Concepts:
- Subatomic particles (protons, neutrons, electrons)
- Charge and mass of subatomic particles

Memory Aids:
- Proton = Positive
- Neutron = Neutral
- Electron = Negative

Quick Tips:
- Remember the basic properties of protons, neutrons, and electrons.

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Application +4 | -1

The energy levels for

{\;_z}{A^{\left( { + z - 1} \right)}}

can be given by:

View answer and explanation

Correct option: A

{E_n}\,for\,\,{A^{\left( { + Z - 1} \right)}} = {Z^2} \times {E_n}\,for\,\,H

Option A is correct because the energy levels for a hydrogen-like species are proportional to Z squared times the energy levels of hydrogen.

Solution approach

Correct Answer + Core Reason: ${E_n}\,for\,\,{A^{\left( { + Z - 1} \right)}} = {Z^2} \times {E_n}\,for\,\,H$ . The energy of an electron in a hydrogen-like atom is proportional to $Z^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the energy of an electron in the nth orbit of a hydrogen-like atom:
$E_n = -13.6 \frac{Z^2}{n^2} eV$ \nwhere Z is the atomic number.

Step 2: The energy of the electron in the nth orbit of a hydrogen atom is:
$E_n(H) = -13.6 \frac{1^2}{n^2} eV$

Step 3: The energy of the electron in the nth orbit of the given species is:
$E_n(A^{+Z-1}) = -13.6 \frac{Z^2}{n^2} eV$

Step 4: Comparing the two energies, we get:
$E_n(A^{+Z-1}) = Z^2 E_n(H)$

Why other options are wrong:
- Option B: This is incorrect because the energy is proportional to $Z^2$ , not $Z$ .
- Option C: This is incorrect because the energy is proportional to $Z^2$ , not $\frac{1}{Z^2}$ .
- Option D: This is incorrect because the energy is proportional to $Z^2$ , not $\frac{1}{Z}$ .

Key Concepts:
- Bohr's model of the atom
- Energy levels of hydrogen-like species

Memory Aids:
- Remember that energy is proportional to $Z^2$ .

Quick Tips:
- Always remember the formula for energy levels in hydrogen-like atoms.

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Which of the following statement(s) is/are incorrect regarding photoelectric effect?

(i) The number of electrons ejected is proportional to the intensity of light.

(ii) There is some time lag between the striking of light beam on the metal surface and the ejection of electrons.

(iii) The kinetic energy of ejected electrons depends upon the brightness of light.

(iv) The kinetic energy of the ejected electron is proportional to the frequency of the incident radiation.

View answer and explanation

Correct option: B

(ii) and (iii)

Option B is correct. There is no time lag in the photoelectric effect, and the kinetic energy depends on the frequency, not the brightness, of the light.

Solution approach

Correct Answer + Core Reason: (ii) and (iii). There is no time lag in the photoelectric effect, and kinetic energy depends on frequency, not brightness.

Step-by-Step Solution:

Step 1: Analyze statement (i): The number of electrons ejected is proportional to the intensity of light. This is correct.

Step 2: Analyze statement (ii): There is some time lag between the striking of light beam on the metal surface and the ejection of electrons. This is incorrect; the process is instantaneous.

Step 3: Analyze statement (iii): The kinetic energy of ejected electrons depends upon the brightness of light. This is incorrect; kinetic energy depends on the frequency of light.

Step 4: Analyze statement (iv): The kinetic energy of the ejected electron is proportional to the frequency of the incident radiation. This is correct.

Step 5: Identify incorrect statements: Statements (ii) and (iii) are incorrect.

Why other options are wrong:
- Option A: (i) and (ii): Incorrect because statement (i) is correct.
- Option C: (ii) only: Incorrect because statement (iii) is also incorrect.
- Option D: (ii) and (iv): Incorrect because statement (iv) is correct.

Key Concepts:
- Photoelectric effect
- Intensity of light
- Frequency of light
- Kinetic energy of electrons

Memory Aids:
- Intensity relates to the number of electrons, frequency to their kinetic energy.

Quick Tips:
- Remember that the photoelectric effect is instantaneous and that kinetic energy depends on frequency.

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Wavelengths of different radiations are given below, arrange in increasing order of energies

\lambda \left( A \right) = 300\,nm,\,\lambda \left( B \right) = 30\mu m,\,\lambda \left( C \right) = 3\,nm,\,\lambda \left( D \right) = 30{A^ \circ }

View answer and explanation

Correct option: A

B < A < C = D

Option A (

B < A < C = D

) is correct because energy is inversely proportional to wavelength. Shorter wavelengths correspond to higher energy.

Solution approach

Correct Answer + Core Reason: $B < A < C = D$ . Energy is inversely proportional to wavelength.

Step-by-Step Solution:

Step 1: Recall the relationship between energy (E) and wavelength ( $\lambda$ ): $E = \frac{{hc}}{\lambda}$ , where h is Planck's constant and c is the speed of light. This means energy is inversely proportional to wavelength.

Step 2: Convert all wavelengths to the same unit (meters):
- $\lambda(A) = 300 \text{{ nm}} = 300 \times 10^{-9} \text{{ m}}$
- $\lambda(B) = 30 \mu \text{{m}} = 30 \times 10^{-6} \text{{ m}}$
- $\lambda(C) = 3 \text{{ nm}} = 3 \times 10^{-9} \text{{ m}}$
- $\lambda(D) = 30 \text{{ Å}} = 30 \times 10^{-10} \text{{ m}} = 3 \times 10^{-9} \text{{ m}}$

Step 3: Compare the wavelengths:
- $\lambda(B) = 30 \times 10^{-6} \text{{ m}}$ (longest wavelength, lowest energy)
- $\lambda(A) = 300 \times 10^{-9} \text{{ m}}$
- $\lambda(C) = 3 \times 10^{-9} \text{{ m}}$
- $\lambda(D) = 3 \times 10^{-9} \text{{ m}}$ (shortest wavelength, highest energy)

Step 4: Arrange in increasing order of energy (decreasing order of wavelength): $B < A < C = D$

Why other options are wrong:
- **Option B ( $A < B < C < D$ ): Incorrect order. It places A before B, implying A has lower energy than B, which is wrong.
- Option C ( $B < A < C < D$ ): Incorrect because C and D have the same wavelength and therefore the same energy.
- Option D ( $B < C < A < D$ ): Incorrect order. It places C before A and incorrectly distinguishes between C and D.

Key Concepts:**
- Relationship between energy and wavelength: $E = \frac{{hc}}{\lambda}$
- Unit conversions (nm, μm, Å to meters)

Memory Aids:
- Remember that shorter wavelengths have higher energy (inverse relationship).

Quick Tips:
- Always convert all wavelengths to the same unit before comparing.
- Remember the electromagnetic spectrum and the relative energies of different types of radiation.

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Compared to mass of lightest nucleus the mass of an electron is only:

View answer and explanation

Correct option: C

1/1800

Option C is correct because the mass of an electron is approximately 1/1800th the mass of the lightest nucleus (Hydrogen).

Solution approach

Correct Answer + Core Reason: $1/1800$ . The mass of an electron is approximately 1/1800th the mass of the lightest nucleus (Hydrogen).

Step-by-Step Solution:

Step 1: Recall the mass of an electron and the mass of a proton (approximate mass of the lightest nucleus).

Step 2: Compare the masses.

Step 3: The mass of an electron is approximately 1/1800th the mass of a proton.

Why other options are wrong:
- Option A: $1/80$ is incorrect.
- Option B: $1/360$ is incorrect.
- Option D: $1/1000$ is incorrect.

Key Concepts:
- Mass of an electron
- Mass of a proton

Memory Aids:
- Electron is much lighter than a proton.

Quick Tips:
- Remember the relative masses of electrons and protons.

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The three quantum numbers  

n,l

and

m

 are the outcome of:

View answer and explanation

Correct option: B

Solution of Schrödinger principle

Option B (Solution of Schr\"odinger principle) is correct because the three quantum numbers

n, l,

and

m

arise from solving the Schr\"odinger equation for the hydrogen atom.

Solution approach

Correct Answer + Core Reason: Solution of Schr\\"odinger principle. The three quantum numbers are the outcome of solving the Schr\\"odinger equation.

Step-by-Step Solution:

Step 1: The question asks about the origin of the three quantum numbers $n, l,$ and $m$ .

Step 2: Recall the origin of quantum numbers:
- The quantum numbers arise from solving the Schr\\"odinger equation for the hydrogen atom.
- $n$ (principal quantum number) determines the energy level.
- $l$ (azimuthal quantum number) determines the shape of the orbital.
- $m$ (magnetic quantum number) determines the orientation of the orbital in space.

Step 3: Identify the correct origin.

Step 4: The solution of the Schr\\"odinger equation is the correct answer.

Why other options are wrong:
- Option A (Bohr’s atomic theory): This is incorrect. Bohr's theory introduced the concept of quantized energy levels but did not fully explain the origin of all three quantum numbers.
- Option C (Heisenberg’s uncertainty principle): This is incorrect. The uncertainty principle deals with the limits of knowing position and momentum simultaneously.
- **Option D ( ${\mathrm{Aufbau}}$  principle): This is incorrect. The Aufbau principle describes the filling of electrons into orbitals.

Key Concepts:
- Quantum numbers
- Schr\\"odinger equation
- Atomic orbitals

Memory Aids:
- Remember that the Schr\\"odinger equation is the foundation for understanding atomic structure and quantum numbers.

Quick Tips:**
- Understand the connection between the Schr\\"odinger equation and the quantum numbers.

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The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency '

\mathrm{A}

'

\times 10^{12}

hertz and that has a radiant intensity in that direction of

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

watt per steradian. 'A' and 'B' are respectively

View answer and explanation

Correct option: B

540 and 683

Option B (540 and 683) is correct because the candela is defined based on a monochromatic radiation of frequency 540 × 10¹² Hz and a radiant intensity of 1/683 watt per steradian.

Solution approach

Correct Answer + Core Reason: 540 and 683. These are the standard values in the definition of candela.

Step-by-Step Solution:

Step 1: Recall the definition of candela.

Step 2: The candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 10¹² hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

Step 3: Identify the values of A and B.

Step 4: A = 540 and B = 683.

Why other options are wrong:
- Option A: The value for B is incorrect.
- Option C: The value for A is incorrect.
- Option D: The value for A is incorrect.

Key Concepts:
- Candela
- Luminous intensity
- SI units

Memory Aids:
- Remember the definition of candela.

Quick Tips:
- Know the definitions of SI base units.

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Which one of the following is the set of correct quantum numbers of an electron in 3d orbital?

View answer and explanation

Correct option: D

n = 3,\;l = 2,\;m = 1,\;s = + 1/2

Option D (

n = 3,\;l = 2,\;m = 1,\;s = + 1/2

) is correct because it provides a valid set of quantum numbers for an electron in a 3d orbital: n=3 (3rd energy level), l=2 (d-orbital), m=1 (possible magnetic quantum number for d-orbital), and s=+1/2 (possible spin quantum number).

Solution approach

Correct Answer + Core Reason: $n = 3,\;l = 2,\;m = 1,\;s = + 1/2$ . This is a valid set of quantum numbers for a 3d electron.

Step-by-Step Solution:

Step 1: Analyze each option based on the rules for quantum numbers.

Step 2: Option A: $n = 3,\;l = 0,\;m = 0,\;s = - 1/2$ . l=0 corresponds to an s-orbital, not a d-orbital.

Step 3: Option B: $n = 2,\;l = 3,\;m = 0,\;s = + 1/2$ . l must be less than n. Here, l=3 and n=2, which is not possible.

Step 4: Option C: $n = 3,\;l = 1,\;m = 0,\;s = - 1/2$ . l=1 corresponds to a p-orbital, not a d-orbital.

Step 5: Option D: $n = 3,\;l = 2,\;m = 1,\;s = + 1/2$ . n=3, l=2 (d-orbital), m can be -2, -1, 0, 1, 2, and s can be +1/2 or -1/2. This is a valid set of quantum numbers.

Why other options are wrong:
- Option A: $n = 3,\;l = 0,\;m = 0,\;s = - 1/2$ . Incorrect because l=0 represents an s-orbital, not a d-orbital.
- Option B: $n = 2,\;l = 3,\;m = 0,\;s = + 1/2$ . Incorrect because l must be less than n.
- Option C: $n = 3,\;l = 1,\;m = 0,\;s = - 1/2$ . Incorrect because l=1 represents a p-orbital, not a d-orbital.

Key Concepts:
- Quantum Numbers
- Atomic Orbitals

Memory Aids:
- Remember the rules for quantum numbers: n>0, 0 <= l < n, -l <= m <= +l, s = +1/2 or -1/2

Quick Tips:
- Always check if the quantum numbers satisfy the required conditions.

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Understanding +4 | -1

Which electronic level will allow the hydrogen atom to absorb photon but not to emit?

View answer and explanation

Correct option: A

1s

Option A (1s) is correct because an electron in the 1s orbital is in the ground state and can only absorb a photon to move to a higher energy level.

Solution approach

Correct Answer + Core Reason: 1s. An electron in the 1s orbital can only absorb a photon, not emit.

Step-by-Step Solution:

Step 1: Define absorption and emission: Absorption is the process where an atom gains energy by absorbing a photon and transitions to a higher energy level. Emission is the process where an atom loses energy by emitting a photon and transitions to a lower energy level.

Step 2: Consider the 1s orbital: The 1s orbital is the ground state of the hydrogen atom, which is the lowest energy level. An electron in the 1s orbital can only absorb a photon to move to a higher energy level.

Step 3: Consider the other orbitals: Electrons in 2s, 2p, and 2d orbitals are in excited states and can either absorb or emit photons.

Step 4: Final Answer: 1s

Why other options are wrong:
- Option B: 2s: An electron in the 2s orbital can both absorb and emit photons.
- Option C: 2p: An electron in the 2p orbital can both absorb and emit photons.
- Option D: 2d: The 2d orbital does not exist. For n=2, l can only be 0 or 1 (s or p).

Key Concepts:
- Atomic energy levels
- Absorption and emission of photons
- Ground state
- Excited states

Memory Aids:
- Ground state (1s) can only absorb.

Quick Tips:
- Remember that the ground state is the lowest energy level, and an electron in the ground state can only absorb a photon.

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Analyzing +4 | -1

Assertion: The boundary surface for

|\psi|^2

and

|\psi|

are identical.

Reason : From the value of

|\psi|^2

at different points within an atom, it is not possible to predict the region around the nucleus where electron will most probably be found.

View answer and explanation

Correct option: C

If A is true but R is false

Option C is correct because the assertion is true, but the reason is false. The boundary surfaces are indeed identical, but the reason provided about predicting electron location is incorrect.

Solution approach

Correct Answer + Core Reason: If A is true but R is false. The boundary surface depends on the angular part of the wavefunction, and $|\psi|^2$ and $|\psi|$ have the same angular dependence. The reason incorrectly states that we cannot predict the region where the electron will most probably be found, which is false; $|\psi|^2$ gives the probability density.

Step-by-Step Solution:

Step 1: Understand the assertion: The boundary surface diagrams represent the shape of the orbitals, which are determined by the angular part of the wavefunction. The boundary surface for both $|\psi|^2$ and $|\psi|$ will be the same because the shape is determined by the angular part, and squaring the wavefunction only affects the radial distribution, not the shape.

Step 2: Understand the reason: The reason states that it's impossible to predict the region around the nucleus where the electron will most probably be found from $|\psi|^2$ . This is incorrect. $|\psi|^2$ represents the probability density of finding an electron at a particular point in space. Therefore, regions with higher $|\psi|^2$ values are more likely to contain the electron.

Step 3: Evaluate the assertion and reason: The assertion is true, as the boundary surfaces for both $|\psi|^2$ and $|\psi|$ are identical. The reason is false because $|\psi|^2$ directly provides information about the probability of finding an electron in a specific region.

Step 4: Conclusion: Since the assertion is true and the reason is false, the correct answer is C.

Why other options are wrong:
- Option A: This is incorrect because the reason is false.
- Option B: This is incorrect because the reason is false.
- Option D: This is incorrect because the assertion is true.

Key Concepts:
- Wavefunction ( $\psi$ )
- Probability density ( $|\psi|^2$ )
- Boundary surface diagrams
- Atomic orbitals

Memory Aids:
- $|\psi|^2$ is the probability density.

Quick Tips:
- Remember that $|\psi|^2$ gives the probability of finding an electron at a point.

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Understanding +4 | -1

Which of the following statement is not correct about the characteristics of cathode rays.

View answer and explanation

Correct option: D

Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube

Option D (Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube) is not correct because the characteristics of cathode rays depend only on the accelerating voltage and not on the gas.

Solution approach

Correct Answer + Core Reason: Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube. Cathode rays are electrons, and their properties do not depend on the gas.

Step-by-Step Solution:

Step 1: Recall the properties of cathode rays.
- Cathode rays are streams of electrons.
- They travel in straight lines.
- They are deflected by electric and magnetic fields.
- Their characteristics (charge-to-mass ratio) do not depend on the material of the electrodes or the gas in the tube.

Step 2: Evaluate each statement:
- A. They start from cathode and move towards the anode - Correct.
- B. They travel in straight line in the absence of an external electrical or magnetic field - Correct.
- C. Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube - Correct.
- D. Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube - Incorrect.

Why other options are wrong:
- Option A: They start from cathode and move towards the anode - This is a correct statement.
- Option B: They travel in straight line in the absence of an external electrical or magnetic field - This is a correct statement.
- Option C: Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube - This is a correct statement.

Key Concepts:
- Cathode rays
- Properties of electrons

Memory Aids:
- Remember that cathode rays are electrons.

Quick Tips:
- Understand the properties of cathode rays and how they were discovered.

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Analyzing +4 | -1

If

{E_e},{E_\alpha}\,

and

{E_p}

represent the kinetic energies of an electron, alpha particle and a proton respectively, each moving with same de-Broglie wavelength then

View answer and explanation

Correct option: D

{E_e} > {E_p} > {E_\alpha }

Option D (

{E_e} > {E_p} > {E_\alpha }

) is correct because for the same de Broglie wavelength, kinetic energy is inversely proportional to mass, and the electron has the smallest mass, followed by the proton, and then the alpha particle.

Solution approach

Correct Answer + Core Reason: ${E_e} > {E_p} > {E_\alpha }$ . For the same de-Broglie wavelength, kinetic energy is inversely proportional to mass.

Step-by-Step Solution:

Step 1: Define de Broglie wavelength: $\lambda = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is momentum.

Step 2: Relate momentum and kinetic energy: $p = \sqrt{2mE}$ , where $m$ is mass and $E$ is kinetic energy.

Step 3: Substitute into the de Broglie wavelength equation: $\lambda = \frac{h}{\sqrt{2mE}}$

Step 4: Rearrange to solve for kinetic energy: $E = \frac{h^2}{2m\lambda^2}$

Step 5: Since the de Broglie wavelength is the same for all particles, $E \propto \frac{1}{m}$

Step 6: The masses are related as $m_e < m_p < m_\alpha$ (mass of electron < mass of proton < mass of alpha particle).

Step 7: Therefore, the kinetic energies are related as $E_e > E_p > E_\alpha$ .

Step 8: Final Answer: ${E_e} > {E_p} > {E_\alpha }$

Why other options are wrong:
- Option A: This is incorrect as it implies the kinetic energies are equal.
- Option B: This is incorrect as it has the wrong order of kinetic energies.
- Option C: This is incorrect as it has the wrong order of kinetic energies.

Key Concepts:
- de Broglie wavelength
- Kinetic energy and momentum relationship
- Mass comparison of electron, proton, and alpha particle

Memory Aids:
- Remember the inverse relationship between kinetic energy and mass for the same de Broglie wavelength.

Quick Tips:
- Understand the relationship between wavelength, momentum, and kinetic energy.
- Remember the relative masses of electron, proton, and alpha particle.

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Understanding +4 | -1

The mass of electron moving with velocity of light is:

View answer and explanation

Correct option: C

Infinite

Option C (Infinite) is correct because, according to the theory of relativity, as an object approaches the speed of light, its mass increases towards infinity.

Solution approach

Correct Answer + Core Reason: Infinite. Mass increases towards infinity as velocity approaches the speed of light.

Step-by-Step Solution:

Step 1: Recall the relativistic mass equation: $m = \frac{{m_0}}{{\sqrt{{1 - \frac{{v^2}}{{c^2}}}}}}$
- where $m$ is the relativistic mass, $m_0$ is the rest mass, $v$ is the velocity of the object, and $c$ is the speed of light.

Step 2: Consider the case when $v$ approaches $c$ .
- As $v \rightarrow c$ , $\frac{{v^2}}{{c^2}} \rightarrow 1$ .

Step 3: Substitute $v = c$ into the equation:
- $m = \frac{{m_0}}{{\sqrt{{1 - 1}}}} = \frac{{m_0}}{0} = \infty$

Step 4: Therefore, the mass of an electron moving with the velocity of light approaches infinity.

Why other options are wrong:
- Option A: $2{m_e}$ is incorrect; the mass approaches infinity, not twice the rest mass.
- Option B: $3{m_e}$ is incorrect; the mass approaches infinity, not three times the rest mass.
- Option D: Zero is incorrect; the mass increases, not decreases to zero.

Key Concepts:
- Relativistic mass
- Speed of light
- Theory of relativity

Memory Aids:
- As velocity approaches the speed of light, mass approaches infinity.

Quick Tips:
- Remember the relativistic mass equation and its implications.

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Application +4 | -1

If the wavelength of an electromagnetic radiation is 2000Å, what is its energy in ergs?

View answer and explanation

Correct option: A

9.94 \times {10^{ - 12}}

Option A is correct. The energy is calculated using E = hc/λ, converting the result to ergs, which gives approximately 9.94 × 10^-12 ergs.

Solution approach

Correct Answer + Core Reason: $9.94 \times {10^{ - 12}}$ . The energy is calculated using $E = \frac{hc}{\lambda}$ and converted to ergs.

Step-by-Step Solution:

Step 1: Recall the formula for energy: $E = \frac{hc}{\lambda}$ , where $h = 6.626 \times 10^{-34} \text{ Js}$ , $c = 3 \times 10^8 \text{ m/s}$ , and $\lambda = 2000 \text{ Å} = 2000 \times 10^{-10} \text{ m} = 2 \times 10^{-7} \text{ m}$ .

Step 2: Substitute the values: $E = \frac{(6.626 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{2 \times 10^{-7} \text{ m}} = \frac{19.878 \times 10^{-26}}{2 \times 10^{-7}} \text{ J} = 9.939 \times 10^{-19} \text{ J}$ .

Step 3: Convert Joules to ergs: $E = 9.939 \times 10^{-19} \text{ J} \times \frac{10^7 \text{ ergs}}{1 \text{ J}} = 9.939 \times 10^{-12} \text{ ergs} \approx 9.94 \times 10^{-12} \text{ ergs}$ .

Why other options are wrong:
- Option B: $9.94 \times {10^{ - 19}}$ : Incorrect; this is the energy in Joules, not ergs.
- Option C: $4.97 \times {10^{ - 12}}$ : Incorrect; this is half the correct energy value, likely a calculation error.
- Option D: $4.97 \times {10^{ - 19}}$ : Incorrect; this is half the correct energy value in Joules, likely a calculation error.

Key Concepts:
- Energy of electromagnetic radiation
- Wavelength
- Planck's constant
- Speed of light

Memory Aids:
- Remember the formula $E = \frac{hc}{\lambda}$ and the conversion between Joules and ergs.

Quick Tips:
- Ensure units are consistent before calculating. Convert to ergs at the end if needed.

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Analyzing +4 | -1

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A: Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium.
Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: A

Both A and R are true and R is the correct explanation of A.

Option A (Both A and R are true and R is the correct explanation of A) is correct because the energy of an orbital decreases with increasing atomic number due to the increased nuclear attraction.

Solution approach

Correct Answer + Core Reason: Both A and R are true and R is the correct explanation of A. The energy of an orbital decreases with increasing atomic number.

Step-by-Step Solution:

Step 1: Assertion A: Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium. This is true. The energy of an orbital decreases as the nuclear charge increases.

Step 2: Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number. This is also true. As the atomic number increases, the nuclear charge increases, which results in a stronger attraction between the nucleus and the electrons, thus lowering the energy of the orbitals.

Step 3: The reason R correctly explains assertion A because lithium has a higher atomic number (Z=3) than hydrogen (Z=1), leading to a lower energy for its 2s orbital compared to hydrogen's 2s orbital.

Step 4: Final Answer: Both A and R are true and R is the correct explanation of A.

Why other options are wrong:
- Option B (Both A and R are true but R is NOT the correct explanation of A): R is the correct explanation of A.
- Option C (A is true but R is false): R is true.
- Option D (A is false but R is true): A is true.

Key Concepts:
- Atomic Number
- Nuclear Charge
- Orbital Energy
- Effective Nuclear Charge

Memory Aids:
- Remember that increasing nuclear charge lowers the energy of orbitals.

Quick Tips:
- Consider the effect of nuclear charge when comparing orbital energies of different atoms.

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Remembering +4 | -1

According to Bohr’s theory, angular momentum of electron in N – shell is

View answer and explanation

Correct option: C

\frac{{2h}}{\pi }

Option C (

\frac{{2h}}{\pi }

) is correct because, according to Bohr's theory, the angular momentum of an electron is quantized and given by

L = n\frac{h}{{2\pi}}

, where n is the principal quantum number.

Solution approach

Correct Answer + Core Reason: $\frac{{2h}}{\pi }$ . Angular momentum of electron in N-shell is calculated using Bohr's theory.

Step-by-Step Solution:

Step 1: Define Bohr's theory for angular momentum: According to Bohr's theory, the angular momentum (L) of an electron is quantized and given by the formula: $L = n \frac{h}{{2\pi}}$ , where n is the principal quantum number and h is Planck's constant.

Step 2: Identify the N-shell: The N-shell corresponds to n = 4.

Step 3: Substitute n = 4 into the formula: $L = 4 \frac{h}{{2\pi}} = \frac{{2h}}{{\pi}}$

Step 4: Final Answer: The angular momentum of an electron in the N-shell is $\frac{{2h}}{{\pi}}$ .

Why other options are wrong:
- Option A: This corresponds to n=1, which is the K-shell.
- Option B: This corresponds to n=2, which is the L-shell.
- Option D: This is an incorrect value and does not correspond to any shell.

Key Concepts:
- Bohr's theory
- Angular momentum quantization
- Principal quantum number (n)
- Electron shells (K, L, M, N...)

Memory Aids:
- Remember the formula: $L = n \frac{h}{{2\pi}}$
- K(1), L(2), M(3), N(4)... alphabetical order corresponds to increasing n.

Quick Tips:
- Always identify the correct value of n for the given shell before applying the formula.

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Application +4 | -1

The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in

\mathrm{H}{\mathrm{e}^ + }

would be:

View answer and explanation

Correct option: C

0.1060 nm

Option C (0.1060 nm) is correct because the radius of the nth Bohr orbit is given by

r = \frac{n^2}{Z} * a_0

, where

a_0

is the Bohr radius (0.053 nm). For He+ (Z=2) and n=2,

r = \frac{2^2}{2} * 0.053 nm = 0.106 nm

.

Solution approach

Correct Answer + Core Reason: 0.1060 nm. The radius of the nth Bohr orbit is given by $r = \frac{n^2}{Z} * a_0$ , where $a_0$ is the Bohr radius (0.053 nm).

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit for a hydrogen-like species:
$r_n = \frac{n^2}{Z} a_0$ , where:
$r_n$ is the radius of the nth orbit\nn is the principal quantum number (orbit number)\nZ is the atomic number
$a_0$ is the Bohr radius (0.053 nm)

Step 2: For $\mathrm{H}{\mathrm{e}^ + }$ , Z = 2. The problem asks for the radius of the second orbit, so n = 2.

Step 3: Substitute n = 2, Z = 2, and $a_0$ = 0.053 nm into the equation:
$r_2 = \frac{2^2}{2} \times 0.053 \;\mathrm{nm} = \frac{4}{2} \times 0.053 \;\mathrm{nm} = 2 \times 0.053 \;\mathrm{nm} = 0.106 \;\mathrm{nm}$

Why other options are wrong:
- Option A: Incorrect. This would be the radius of the first orbit of He+ (n=1, Z=2).
- Option B: Incorrect. This is the radius of the first Bohr orbit of hydrogen (n=1, Z=1).
- Option D: Incorrect. This is twice the correct radius of the second orbit of He+.

Key Concepts:
- Bohr's Theory
- Bohr Radius
- Hydrogen-like Species

Memory Aids:
- Remember the formula for the radius of the nth Bohr orbit for hydrogen-like species: $r_n = \frac{n^2}{Z} a_0$ .

Quick Tips:
- Always consider the atomic number (Z) when calculating the Bohr radius for ions.

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Application +4 | -1

The radius of which of the following orbit is same as that of first Bohr’s orbit of Hydrogen atom

View answer and explanation

Correct option: D

B{e^{ + 3}}\left( {n = 2} \right)

Option D (

B{e^{ + 3}} {n = 2}

) is correct because the radius of the orbit is proportional to

n^2/Z

, and for it to be equal to the first Bohr's orbit of Hydrogen,

n^2/Z

must be equal to 1.

Solution approach

Correct Answer + Core Reason: $B{e^{ + 3}}\left( {n = 2} \right)$ . The radius of the orbit is proportional to $\frac{{{n^2}}}{Z}$ , where n is the principal quantum number and Z is the atomic number.

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r = {r_0}\frac{{{n^2}}}{Z}$ , where ${r_0}$ is the Bohr radius for hydrogen (n=1, Z=1).

Step 2: We want to find an ion with an orbit radius equal to the first Bohr orbit of hydrogen, so we need $\frac{{{n^2}}}{Z} = 1$ .

Step 3: Check each option:
- A. $H{e^ + }\left( {n = 2} \right)$ : $Z = 2$ , $n = 2$ , so $\frac{{{n^2}}}{Z} = \frac{{{2^2}}}{2} = 2 \ne 1$ .
- B. $L{i^{ + 2}}\left( {n = 2} \right)$ : $Z = 3$ , $n = 2$ , so $\frac{{{n^2}}}{Z} = \frac{{{2^2}}}{3} = \frac{4}{3} \ne 1$ .
- C. $L{i^{ + 2}}\left( {n = 3} \right)$ : $Z = 3$ , $n = 3$ , so $\frac{{{n^2}}}{Z} = \frac{{{3^2}}}{3} = 3 \ne 1$ .
- D. $B{e^{ + 3}}\left( {n = 2} \right)$ : $Z = 4$ , $n = 2$ , so $\frac{{{n^2}}}{Z} = \frac{{{2^2}}}{4} = 1$ .

Step 4: Since option D satisfies the condition $\frac{{{n^2}}}{Z} = 1$ , the radius of the orbit is the same as the first Bohr's orbit of Hydrogen atom.

Why other options are wrong:
- Option A: $H{e^ + }\left( {n = 2} \right)$ is incorrect because $\frac{{{n^2}}}{Z} = 2 \ne 1$ .
- Option B: $L{i^{ + 2}}\left( {n = 2} \right)$ is incorrect because $\frac{{{n^2}}}{Z} = \frac{4}{3} \ne 1$ .
- Option C: $L{i^{ + 2}}\left( {n = 3} \right)$ is incorrect because $\frac{{{n^2}}}{Z} = 3 \ne 1$ .

Key Concepts:
- Bohr's model of the atom
- Radius of Bohr orbit: $r = {r_0}\frac{{{n^2}}}{Z}$
- Atomic number (Z)
- Principal quantum number (n)

Memory Aids:
- Remember that the radius is proportional to n squared over Z.

Quick Tips:
- Always remember the relationship between radius, principal quantum number, and atomic number.
- Check each option by calculating $\frac{{{n^2}}}{Z}$ .

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Analyzing +4 | -1

Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is :

View answer and explanation

Correct option: B

n = 3\;\mathrm{to\;}n = 8

Option B (

n = 3\;\mathrm{to\;}n = 8

) is correct because the lowest frequency absorption line corresponds to the smallest energy difference between energy levels.

Solution approach

Correct Answer + Core Reason: $n = 3\;\mathrm{to\;}n = 8$ . The absorption line of lowest frequency corresponds to the smallest energy difference between the energy levels.

Step-by-Step Solution:

Step 1: Identify that absorption lines occur when an electron transitions from a lower energy level to a higher energy level.

Step 2: Recall that the energy difference between energy levels is given by $\Delta E = E_f - E_i = -13.6 eV (\frac{1}{n_f^2} - \frac{1}{n_i^2})$ , where $n_i$ is the initial energy level and $n_f$ is the final energy level.

Step 3: Analyze each option to determine the energy difference:
- A: $n = 1 \;\mathrm{to\;} n = 2$ (Absorption)
- B: $n = 3 \;\mathrm{to\;} n = 8$ (Absorption)
- C: $n = 2 \;\mathrm{to\;} n = 1$ (Emission)
- D: $n = 8 \;\mathrm{to\;} n = 3$ (Emission)

Step 4: Since we are looking for absorption, we can eliminate options C and D.

Step 5: Compare the energy differences for options A and B. The energy difference is smaller when both n values are larger. Therefore, the transition from n=3 to n=8 will have a smaller energy difference than n=1 to n=2.

Step 6: Since frequency is directly proportional to energy ( $E = hf$ ), the transition with the smallest energy difference will have the lowest frequency.

Step 7: Conclusion: The transition from $n = 3$ to $n = 8$ gives the absorption line of lowest frequency.

Why other options are wrong:
- Option A: $n = 1\;\mathrm{to\;}n = 2$ requires a larger energy difference than $n = 3\;\mathrm{to\;}n = 8$ , hence higher frequency.
- Option C: $n = 2\;\mathrm{to\;}n = 1$ is an emission, not an absorption.
- Option D: $n = 8\;\mathrm{to\;}n = 3$ is an emission, not an absorption.

Key Concepts:
- Absorption and Emission Spectra
- Energy Levels in Hydrogen Atom
- Rydberg Formula
- Relationship between Energy and Frequency ( $E = hf$ )

Memory Aids:
- Absorption: Electron moves to higher energy level.
- Emission: Electron moves to lower energy level.

Quick Tips:
- Remember that absorption involves transitions to higher energy levels, while emission involves transitions to lower energy levels.
- The smaller the difference in n values at higher n values, the lower the energy and frequency.

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Analyzing +4 | -1

If

{E_1},{E_2}

and

{E_3}

represents kinetic energy of electron,

\alpha -

particle and proton respectively (having same De Broglie wavelength) then

View answer and explanation

Correct option: A

{E_1} > {E_3} > {E_2}

Option A (

{E_1} > {E_3} > {E_2}

) is correct because the kinetic energy is inversely proportional to the mass, and the electron has the smallest mass, followed by the proton, and then the alpha particle.

Solution approach

Correct Answer + Core Reason: ${E_1} > {E_3} > {E_2}$ . Kinetic energy is inversely proportional to mass when the de Broglie wavelength is the same.

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula: $\lambda = \frac{{h}}{{p}}$ , where h is Planck's constant and p is the momentum.

Step 2: Since the de Broglie wavelength is the same for all three particles, their momenta are also the same: $p_1 = p_2 = p_3 = p$ .

Step 3: Recall the relationship between kinetic energy (E) and momentum (p): $E = \frac{{p^2}}{{2m}}$ , where m is the mass of the particle.

Step 4: Since p is the same for all three particles, the kinetic energy is inversely proportional to the mass: $E \propto \frac{{1}}{{m}}$

Step 5: The mass of an electron ( $m_1$ ) is much smaller than the mass of a proton ( $m_3$ ), and the mass of an alpha particle ( $m_2$ ) is approximately four times the mass of a proton. Therefore, $m_1 < m_3 < m_2$ .

Step 6: Since $E \propto \frac{{1}}{{m}}$ , the kinetic energies are related as $E_1 > E_3 > E_2$ .

Why other options are wrong:
- Option B: This option suggests the opposite relationship, where the alpha particle has the highest kinetic energy, which is incorrect because it has the largest mass.
- Option C: This option incorrectly places the kinetic energy of the alpha particle between the electron and the proton. It's a misinterpretation of the inverse relationship between mass and kinetic energy.
- Option D: This option suggests that all the kinetic energies are equal, which is incorrect because the masses are different.

Key Concepts:
- de Broglie wavelength
- Kinetic energy and momentum relationship
- Inverse relationship between kinetic energy and mass (when momentum is constant)

Memory Aids:
- Remember the formula $\lambda = \frac{{h}}{{p}}$ and $E = \frac{{p^2}}{{2m}}$

Quick Tips:
- When comparing kinetic energies with the same de Broglie wavelength, focus on the inverse relationship with mass.
- Remember the relative masses of electron, proton, and alpha particle.

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Remembering +4 | -1

The total energy in

1^{\text {st }}

orbit of hydrogen atom is given by

View answer and explanation

Correct option: D

-\frac{2 \pi^2 k^2 e^4 m}{h^2}

Option D (

-\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.

Solution approach

Correct Answer + Core Reason: $-\frac{2 \pi^2 k^2 e^4 m}{h^2}$ . This is the correct expression for the total energy in the first orbit of a hydrogen atom.

Step-by-Step Solution:

Step 1: Recall the expression for the total energy of an electron in the nth orbit of a hydrogen atom based on Bohr's model.

Step 2: The total energy (E) is given by: $E = -\frac{2 \pi^2 k^2 e^4 m}{n^2 h^2}$ , where k is Coulomb's constant, e is the electron charge, m is the electron mass, n is the principal quantum number, and h is Planck's constant.

Step 3: For the first orbit, n = 1. Therefore, $E_1 = -\frac{2 \pi^2 k^2 e^4 m}{h^2}$ .

Why other options are wrong:
- Option A: $\frac{2 \pi^2 k^2 e^4 m}{h^3}$ is incorrect because the denominator should be $h^2$ , not $h^3$ .
- Option B: $-\frac{2 \pi^2 k^2 e^2 m}{h^2}$ is incorrect because the electron charge should be raised to the power of 4 ( $e^4$ ), not 2 ( $e^2$ ).
- Option C: $\frac{2 \pi^2 k^2 e^4 m}{h^2}$ is incorrect because the total energy should be negative, indicating a bound state.

Key Concepts:
- Bohr's model of the hydrogen atom
- Total energy of an electron in an orbit
- Planck's constant

Memory Aids:
- Remember the formula for the total energy: $E = -\frac{2 \pi^2 k^2 e^4 m}{n^2 h^2}$

Quick Tips:
- Pay attention to the signs and exponents in the formula.

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Understanding +4 | -1

When high speed electrons strike a target:

View answer and explanation

Correct option: D

Heat is produced and simultaneously continuous and characteristic X-rays are emitted

Option D (Heat is produced and simultaneously continuous and characteristic X-rays are emitted) is correct because the collision generates both heat and X-rays.

Solution approach

Correct Answer + Core Reason: Heat is produced and simultaneously continuous and characteristic X-rays are emitted. High-speed electrons striking a target produce both heat and X-rays.

Step-by-Step Solution:

Step 1: Describe the process: When high-speed electrons strike a target material, they decelerate rapidly.

Step 2: Explain heat production: The deceleration of electrons leads to the generation of heat due to collisions with the target atoms.

Step 3: Explain continuous X-ray emission (Bremsstrahlung): The deceleration of electrons also causes them to emit continuous X-rays (Bremsstrahlung) due to the loss of kinetic energy.

Step 4: Explain characteristic X-ray emission: If the electrons have sufficient energy, they can eject core electrons from the target atoms. When these vacancies are filled by other electrons, characteristic X-rays are emitted.

Step 5: Final Answer: Heat is produced and simultaneously continuous and characteristic X-rays are emitted

Why other options are wrong:
- Option A: Only heat is produced: Incorrect, X-rays are also produced.
- Option B: Only continuous X-rays are emitted: Incorrect, heat and characteristic X-rays are also produced.
- Option C: Only continuous and characteristic X-rays are emitted: Incorrect, heat is also produced.

Key Concepts:
- X-ray production
- Bremsstrahlung (continuous X-rays)
- Characteristic X-rays
- Heat generation

Memory Aids:
- High-speed electrons hitting a target = Heat + Continuous X-rays + Characteristic X-rays

Quick Tips:
- Remember that both heat and X-rays are produced when high-speed electrons strike a target.

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Application +4 | -1

Calculate the wavelength (in nanometer) associated with a proton moving at

1.0 \times {10^3}\mathrm{m}{\mathrm{s}^{ - 1}}

(Mass of proton

= 1.67 \times {10^{ - 27}}\mathrm{kg\;and\;}h = 6.63 \times {10^{ - 34}}\mathrm{\;Js}

)

View answer and explanation

Correct option: B

0.40 nm

Option B (0.40 nm) is correct because it's calculated using the de Broglie wavelength formula:

\lambda = \frac{h}{mv}

, where h is Planck's constant, m is mass, and v is velocity.

Solution approach

Correct Answer + Core Reason: 0.40 nm. The wavelength is calculated using the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$ .

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$ , where h is Planck's constant, m is the mass of the particle, and v is its velocity.

Step 2: Given values: $h = 6.63 \times 10^{-34} Js$ , $m = 1.67 \times 10^{-27} kg$ , $v = 1.0 \times 10^3 m/s$ .

Step 3: Substitute the values into the formula: $\lambda = \frac{6.63 \times 10^{-34} Js}{(1.67 \times 10^{-27} kg)(1.0 \times 10^3 m/s)}$ .

Step 4: Calculate the wavelength: $\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-24}} m = 3.97 \times 10^{-10} m$ .

Step 5: Convert the wavelength to nanometers: $\lambda = 3.97 \times 10^{-10} m \times \frac{10^9 nm}{1 m} = 0.397 nm \approx 0.40 nm$ .

Why other options are wrong:

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Remembering +4 | -1

The presence of charge particles in the atoms was first confirmed by

View answer and explanation

Correct option: C

Faraday

Option C (Faraday) is correct because Faraday's laws of electrolysis provided early evidence for the existence of charged particles (ions) in atoms.

Solution approach

Correct Answer + Core Reason: Faraday. Faraday's laws of electrolysis confirmed the presence of charged particles.

Step-by-Step Solution:

Step 1: Recall the contributions of different scientists to the discovery of atomic structure.

Step 2: Faraday's laws of electrolysis demonstrated the relationship between electricity and chemical reactions, suggesting the existence of charged particles (ions).

Why other options are wrong:
- Option A (Rutherford): Rutherford's experiment focused on the nucleus and the scattering of alpha particles.
- Option B (Thomson): Thomson discovered the electron and proposed the plum pudding model.
- Option D (Goldstein): Goldstein discovered canal rays (positive ions), but Faraday's work predates this and provided earlier evidence of charged particles.

Key Concepts:
- Faraday's laws of electrolysis
- Atomic structure

Memory Aids:
- Remember that Faraday's laws of electrolysis provided early evidence for charged particles in atoms.

Quick Tips:
- Know the contributions of different scientists to the development of atomic theory.

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Application +4 | -1

The electronic configuration of an element in ultimate and penultimate orbitals is

\left( {n-1} \right){s^2}\left( {n-1} \right){p^6}\left( {n-1} \right){d^x}\;n{s^2}

. If

n=4

and

x=5

then number of protons in the nucleus is

View answer and explanation

Correct option: C

25

Option C (25) is correct because the electronic configuration given corresponds to Manganese (Mn), which has an atomic number of 25.

Solution approach

Correct Answer + Core Reason: 25. The electronic configuration corresponds to Mn, which has 25 protons.

Step-by-Step Solution:

Step 1: Write out the electronic configuration based on the given information:
- $(n-1)s^2 (n-1)p^6 (n-1)d^x ns^2$
- Given $n = 4$ and $x = 5$ , the configuration becomes: $(4-1)s^2 (4-1)p^6 (4-1)d^5 4s^2$ , which simplifies to $3s^2 3p^6 3d^5 4s^2$ .

Step 2: Write the complete electronic configuration:
- $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$

Step 3: Count the number of electrons:
- $2 + 2 + 6 + 2 + 6 + 5 + 2 = 25$

Step 4: The number of electrons in a neutral atom is equal to the number of protons. Therefore, the number of protons is 25.

Final Answer: 25

Why other options are wrong:
- Option A (>25): This is incorrect because the number of electrons is exactly 25.
- Option B (<24): This is incorrect because the number of electrons is exactly 25.
- Option D (30): This is incorrect. 30 would correspond to Zinc (Zn), which has a different electronic configuration.

Key Concepts:
- Electronic Configuration
- Atomic Number
- Number of Protons and Electrons in a Neutral Atom

Memory Aids:
- Remember the order of filling orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...

Quick Tips:
- Write out the complete electronic configuration to accurately count the number of electrons.

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Analyzing +4 | -1

Select the one among the following transitions in hydrogen atom, which gives an absorption line of lowest frequency

View answer and explanation

Correct option: C

n=4

to

n=6

Option C (

n=4

to

n=6

) is correct because the smallest energy difference between energy levels corresponds to the lowest frequency absorption line.

Solution approach

Correct Answer + Core Reason: $n=4$ to $n=6$ . The smallest energy difference corresponds to the lowest frequency.

Step-by-Step Solution:

Step 1: The frequency of an absorption line is related to the energy difference between the initial and final energy levels by the equation $E = hf$ , where $E$ is the energy difference, $h$ is Planck's constant, and $f$ is the frequency.

Step 2: Therefore, the lowest frequency corresponds to the smallest energy difference. The energy difference between energy levels in a hydrogen atom decreases as n increases. So, transitions between higher energy levels will have smaller energy differences.

Step 3: Analyze the options:
- A. $n=1$ to $n=3$ : This is a large energy difference.
- B. $n=2$ to $n=3$ : This is a smaller energy difference than option A, but still relatively large.
- C. $n=4$ to $n=6$ : This transition occurs between higher energy levels, so the energy difference is smaller.
- D. $n=3$ to $n=4$ : This is a smaller energy difference than option A and B, but still larger than C.

Step 4: To compare C and D more accurately, we can consider the difference in $\frac{1}{n^2}$ for each:
- C: $\frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} = \frac{9-4}{144} = \frac{5}{144}$
- D: $\frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} = \frac{16-9}{144} = \frac{7}{144}$

Step 5: Since $\frac{5}{144} < \frac{7}{144}$ , the transition $n=4$ to $n=6$ has the smallest energy difference and therefore the lowest frequency.

Step 6: Final Answer: $n=4$ to $n=6$

Why other options are wrong:
- **Option A ( $\mathrm{n}=1$ to $\mathrm{n}=3$ ): This transition has a larger energy difference and therefore a higher frequency.
- Option B ( $\mathrm{n}=2$ to $\mathrm{n}=3$ ): This transition has a larger energy difference and therefore a higher frequency.
- Option D ( $\mathrm{n}=3$ to $\mathrm{n}=4$ ): This transition has a larger energy difference compared to option C and therefore a higher frequency.

Key Concepts:
- Hydrogen Atomic Spectra
- Energy Levels
- Electronic Transitions
- Absorption and Emission
- Frequency and Energy

Memory Aids:
- Remember that the energy difference decreases as n increases.

Quick Tips:**
- The transition between higher energy levels will have a smaller energy difference and a lower frequency.

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Remembering +4 | -1

Maxwell revealed that light waves are associated with

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because Maxwell's theory states that light waves are electromagnetic, having both oscillating electric and magnetic components.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). Light waves are electromagnetic waves with oscillating electric and magnetic fields.

Step-by-Step Solution:

Step 1: The question asks about the nature of light waves according to Maxwell's theory.

Step 2: Recall that Maxwell's theory describes light as an electromagnetic wave, which consists of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation.

Step 3: Therefore, light waves are associated with both oscillating electric and magnetic character.

Why other options are wrong:
- Option A: Oscillating electric character is incorrect because it only describes one aspect of light waves. Light waves also have an oscillating magnetic character.
- Option B: Oscillating magnetic character is incorrect because it only describes one aspect of light waves. Light waves also have an oscillating electric character.
- Option D: None of the above is incorrect because light waves are associated with both oscillating electric and magnetic character.

Key Concepts:
- Maxwell's theory of electromagnetism
- Electromagnetic waves

Memory Aids:
- Light is an electromagnetic wave.

Quick Tips:
- Remember that light has both electric and magnetic components.

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Analyzing +4 | -1

Assertion: The transition of electrons

n_3 \rightarrow n_2

in H-atom will emit radiation of higher frequency than

n_4 \rightarrow n_3

.

Reason: Principal shells

n_2

and

n_3

have lower energy than

{n}_4

.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the energy difference between levels is the direct cause of frequency, not just the energy of the levels themselves.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. Both statements are true, but the reason doesn't directly explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The transition of electrons $n_3 \rightarrow n_2$ in H-atom will emit radiation of higher frequency than $n_4 \rightarrow n_3$ .

Step 2: The frequency of emitted radiation is proportional to the energy difference between the levels. The energy difference between $n=3$ and $n=2$ is greater than the energy difference between $n=4$ and $n=3$ . Therefore, the assertion is correct.

Step 3: Analyze the Reason: Principal shells $n_2$ and $n_3$ have lower energy than ${n}_4$ .

Step 4: This statement is also correct. Energy levels increase as n increases, so $n_2$ and $n_3$ have lower energy than $n_4$ .

Step 5: Determine if the Reason is the correct explanation of the Assertion. While both are true, the reason doesn't directly explain why the frequency is higher. The higher frequency is due to the larger energy *difference* between $n_3$ and $n_2$ compared to the energy difference between $n_4$ and $n_3$ .

Why other options are wrong:
- Option A: The Reason is not the correct explanation of the Assertion.
- Option C: The Reason is correct, not incorrect.
- Option D: Both the Assertion and Reason are correct.

Key Concepts:
- Energy Levels in Hydrogen Atom
- Frequency of Emitted Radiation

Memory Aids:
- Higher energy level transitions result in higher frequency radiation.

Quick Tips:
- Consider both the energy levels and the energy differences between levels when analyzing transitions.

60 Medium

60 of 193Question permalink

Remembering +4 | -1

The orbital angular momentum of an electron revolving in a p-otbital is

View answer and explanation

Correct option: B

\frac{h}{{\sqrt {2\pi } }}

Option B (

\frac{h}{{\sqrt {2\pi } }}

) is correct because the orbital angular momentum for a p-orbital (l=1) is calculated using the formula

\sqrt{l(l+1)}\frac{h}{2\pi}

.

Solution approach

Correct Answer + Core Reason: $\frac{h}{{\sqrt {2\pi } }}$ . The orbital angular momentum depends on the azimuthal quantum number (l).

Step-by-Step Solution:

Step 1: Recall the formula for orbital angular momentum.

Step 2: The orbital angular momentum is given by $\sqrt{l(l+1)} \frac{h}{2\pi}$ , where l is the azimuthal quantum number.

Step 3: For a p-orbital, l = 1.

Step 4: Substitute l = 1 into the formula: $\sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} = \frac{h}{\sqrt{2}\pi} = \frac{h}{\sqrt{2\pi^2}} = \frac{h}{\sqrt{2\pi}}$

Why other options are wrong:
- Option A: Zero. This is the orbital angular momentum for an s-orbital (l=0).
- Option C: $\frac{h}{{2\pi }}$ . This would be the angular momentum if the $\sqrt{l(l+1)}$ term was equal to 1, which is not the case for a p-orbital.
- Option D: $\frac{1}{2}\frac{h}{{2\pi }}$ . This is incorrect.

Key Concepts:
- Orbital Angular Momentum
- Azimuthal Quantum Number
- Atomic Orbitals

Memory Aids:
- Remember the formula: $\sqrt{l(l+1)} \frac{h}{2\pi}$

Quick Tips:
- Know the l values for s, p, d, and f orbitals (0, 1, 2, 3 respectively).

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