${CO}_2, {~N}_2 {O}, {NO}_3^{-}$

STRUCTURE OF ATOM MCQs for JEE CHEMISTRY — Easy, Page 7

181 Easy

181 of 239

Understanding +4 | -1

Heisenberg's uncertainity principle is applicable to

View answer and explanation

Correct option: D

Any moving object

Option D (Any moving object) is correct because Heisenberg's uncertainty principle applies to all moving objects, not just atoms, electrons, or nuclei.

Solution approach

Correct Answer + Core Reason: Any moving object. Heisenberg's uncertainty principle is a fundamental concept applicable to all moving objects.

Step-by-Step Solution:

Step 1: Heisenberg's uncertainty principle states that it is impossible to determine simultaneously both the position and momentum of a particle with perfect accuracy.

Step 2: This principle is a consequence of wave-particle duality and applies to all objects, regardless of size or mass.

Step 3: The mathematical formulation of the principle is given by: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$ , where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant.

Step 4: This principle is not limited to atoms, electrons, or nuclei; it applies to any moving object.

Why other options are wrong:
- Option A: Atoms only - This is incorrect because the principle applies to all moving objects, not just atoms.
- Option B: Electron only - This is incorrect because the principle applies to all moving objects, not just electrons.
- Option C: Nucleus only - This is incorrect because the principle applies to all moving objects, not just nuclei.
- Option D: Any moving object - This is correct because the principle is a fundamental aspect of quantum mechanics and applies universally.

Key Concepts:
- Heisenberg's Uncertainty Principle
- Wave-particle duality
- Quantum mechanics

Memory Aids:
- Remember that uncertainty is inherent in measuring position and momentum simultaneously.

Quick Tips:
- The uncertainty principle is more noticeable for small particles like electrons due to their small mass.

182 Easy

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Remembering +4 | -1

Electrons are filled in energy orbitals, in increasing order of energy." This statement is related to

View answer and explanation

Correct option: D

Aufbau principle

Option D (Aufbau principle) is correct because the Aufbau principle states that electrons fill atomic orbitals in order of increasing energy.

Solution approach

Correct Answer + Core Reason: Aufbau principle. The Aufbau principle dictates that electrons fill orbitals in increasing order of energy.

Step-by-Step Solution:

Step 1: Understand the statement. The statement describes how electrons are arranged in atomic orbitals based on their energy levels.

Step 2: Recall the Aufbau principle. The Aufbau principle states that electrons first occupy the lowest energy orbitals available before filling higher energy orbitals.

Step 3: Relate the statement to the Aufbau principle. The statement is a direct description of the Aufbau principle.

Why other options are wrong:
- Option A: Planck's rule relates to the quantization of energy and is not directly related to the filling of electron orbitals.
- Option B: Hund's rule states that electrons individually occupy each orbital within a subshell before any orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. It doesn't describe the overall filling order.
- Option C: Pauli's exclusion principle states that no two electrons in the same atom can have the same set of four quantum numbers. It restricts the number of electrons in an orbital but doesn't define the filling order.
- Option D: This option is correct because the Aufbau principle describes the filling of electrons in increasing order of energy.

Key Concepts:
- Aufbau principle
- Hund's rule
- Pauli's exclusion principle
- Electronic configuration

Memory Aids:
- Aufbau = \"building up,\" referring to the way electrons are added to orbitals.

Quick Tips:
- Remember the order of filling orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.

183 Easy

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Understanding +4 | -1

The outer electronic structure

3 s^2 3 p^5

is possessed by

View answer and explanation

Correct option: A

{Cl}

Option A (

{Cl}

) is correct because its electronic configuration ends with

3s^2 3p^5

.

Solution approach

Correct Answer + Core Reason: ${Cl}$ . Chlorine has an electronic configuration that ends with $3s^2 3p^5$ .

Step-by-Step Solution:

Step 1: Identify the element with the given electronic configuration: The electronic configuration is $3s^2 3p^5$ .

Step 2: Determine the number of electrons: The number of electrons in the outermost shell is 2 + 5 = 7.

Step 3: Identify the element: The element with 7 valence electrons in the 3rd period is chlorine (Cl).

Step 4: Write the complete electronic configuration of chlorine: $1s^2 2s^2 2p^6 3s^2 3p^5$ .

Why other options are wrong:
- **Option B ( $O$ ):** Oxygen has the electronic configuration $1s^2 2s^2 2p^4$ .
- **Option C ( ${Ar}$ ):** Argon has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ .
- **Option D ( ${Br}$ ):** Bromine has the electronic configuration $[Ar] 4s^2 3d^{10} 4p^5$ .

Key Concepts:
- Electronic configuration
- Valence electrons
- Periodic table

Memory Aids:
- Remember the electronic configurations of common elements.

Quick Tips:
- Use the periodic table to quickly identify elements based on their electronic configurations.

184 Easy

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Understanding +4 | -1

In potassium the order of energy level for 19th electron is:

View answer and explanation

Correct option: B

4s<3d

Option B (4s<3d) is correct because the (n+l) rule dictates that the 4s orbital is filled before the 3d orbital due to its lower energy.

Solution approach

Correct Answer + Core Reason: 4s<3d. $\left( {n + l} \right)$ is more for a subshell, more will be its energy.

Step-by-Step Solution:
Step 1: State the (n+l) rule: The (n+l) rule states that electrons first fill the orbital with the lowest value of (n+l), where n is the principal quantum number and l is the azimuthal quantum number.
Step 2: Calculate (n+l) for 4s and 3d: For 4s, n=4 and l=0, so (n+l) = 4. For 3d, n=3 and l=2, so (n+l) = 5.
Step 3: Compare the values: Since 4 < 5, the 4s orbital has lower energy than the 3d orbital.
Step 4: Apply to potassium: In potassium, the 19th electron fills the 4s orbital before the 3d orbital.

Why other options are wrong:
- Option A: 3s>3d is incorrect because 3d has higher energy than 3s.
- Option C: 4s>4p is incorrect because 4p has higher energy than 4s.
- Option D: 4s=3d is incorrect because they have different energies.

Key Concepts:
- Aufbau Principle
- (n+l) Rule
- Electronic Configuration
- Energy Levels of Orbitals

Memory Aids:
- Remember the (n+l) rule for determining the order of filling orbitals.

Quick Tips:
- Use the (n+l) rule to predict the electronic configuration of elements.

185 Easy

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Understanding +4 | -1

For which species, Bohr's theory does not apply:

View answer and explanation

Correct option: B

Be

Option B (Be) is correct because Bohr's theory is only applicable to single-electron systems, and Be has more than one electron.

Solution approach

Correct Answer + Core Reason: Be. Bohr's theory is applicable to one-electron systems only.

Step-by-Step Solution:
Step 1: Recall the postulates of Bohr's theory.
Step 2: Bohr's theory is based on the hydrogen atom, which has only one electron.
Step 3: The theory can be extended to hydrogen-like species, which are ions with only one electron.
Step 4: Check the number of electrons in each species:\nH: 1 electron\nBe: 4 electrons
$\mathrm{He}^+$ : 1 electron
$\mathrm{Li}^{2+}$ : 1 electron
Step 5: Bohr's theory is not applicable to Be because it has more than one electron.

Why other options are wrong:
- Option A: H has only one electron, so Bohr's theory applies.
- Option C: $\mathrm{He}^+$ has only one electron, so Bohr's theory applies.
- Option D: $\mathrm{Li}^{2+}$ has only one electron, so Bohr's theory applies.

Key Concepts:
- Bohr's theory
- Hydrogen-like species

Memory Aids:
- Bohr's model works for single-electron systems only.

Quick Tips:
- Identify the number of electrons in the atom or ion to determine if Bohr's theory is applicable.

186 Easy

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Remembering +4 | -1

Which is the correct relationship between wavelength and momentum of particles ?

View answer and explanation

Correct option: A

\lambda=\frac{h}{p}

Option A (

\lambda=\frac{h}{p}

) is correct because it represents the de Broglie wavelength equation, relating wavelength and momentum.

Solution approach

Correct Answer + Core Reason: $\lambda=\frac{h}{p}$ . This is the de Broglie wavelength equation.

Step-by-Step Solution:

Step 1: Recall the de Broglie equation.

Step 2: The de Broglie equation relates the wavelength of a particle to its momentum: $\lambda = \frac{h}{p}$ , where $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum.

Step 3: Therefore, the correct relationship is $\lambda=\frac{h}{p}$ .

Why other options are wrong:
- Option B: $\pi=\frac{h}{p}$ is incorrect. This equation is not a valid physical relationship.
- Option C: $p=\frac{h}{\lambda m}$ is incorrect. The correct equation is $p = \frac{h}{\lambda}$ .
- Option D: $h=\frac{p}{\lambda}$ is incorrect. The correct equation is $\lambda = \frac{h}{p}$ .

Key Concepts:
- de Broglie Wavelength
- Momentum
- Planck's Constant

Memory Aids:
- $\lambda = h/p$ (Lambda equals h over p)

Quick Tips:
- Remember the de Broglie equation and its components.

187 Easy

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Application +4 | -1

Chlorophyll present in green leaves of plants absorbs light at

4.620 \times {10^{14}}Hz

. Calculate the wavelength of radiation in nm.

View answer and explanation

Correct option: A

649.4

Option A (649.4) is correct because the wavelength is calculated using the formula

\lambda = c/f

, where c is the speed of light and f is the frequency.

Solution approach

Correct Answer + Core Reason: 649.4. The wavelength is calculated using the relationship between speed of light, frequency, and wavelength.

Step-by-Step Solution:

Step 1: Identify the given values: frequency (f) = $4.620 \times 10^{14}$ Hz, speed of light (c) = $3.0 \times 10^8$ m/s.

Step 2: Use the formula: $\lambda = \frac{c}{f}$ , where $\lambda$ is the wavelength.

Step 3: Substitute the values: $\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{4.620 \times 10^{14} \text{ Hz}}$ .

Step 4: Calculate the wavelength in meters: $\lambda = 6.4935 \times 10^{-7} \text{ m}$ .

Step 5: Convert the wavelength to nanometers: $\lambda = 6.4935 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 649.35 \text{ nm}$ .

Step 6: Round to one decimal place: $\lambda \approx 649.4 \text{ nm}$ .

Why other options are wrong:
- Option B (64.94): This is incorrect due to a miscalculation or incorrect unit conversion.
- Option C (6.494): This is incorrect due to a miscalculation or incorrect unit conversion.
- Option D (649.4): This is the correct answer.

Key Concepts:
- Relationship between speed of light, frequency, and wavelength: $c = f\lambda$
- Unit conversion (meters to nanometers)

Memory Aids:
- Remember the formula: $\lambda = \frac{c}{f}$

Quick Tips:
- Always use consistent units.
- Remember the conversion factor between meters and nanometers.

188 Easy

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Application +4 | -1

If the K.E of a particle is reduced to half, de Broglie wavelength becomes

View answer and explanation

Correct option: D

\sqrt 2

times

Option D (

\sqrt 2

times) is correct because de Broglie wavelength is inversely proportional to the square root of kinetic energy.

Solution approach

Correct Answer + Core Reason: $\sqrt 2$ times. The de Broglie wavelength is inversely proportional to the square root of the kinetic energy.

Step-by-Step Solution:

Step 1: Define the de Broglie wavelength formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$ , where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $p$ is momentum, $m$ is mass, and $KE$ is kinetic energy.

Step 2: Analyze the relationship between $\lambda$ and $KE$ : $\lambda \propto \frac{1}{\sqrt{KE}}$

Step 3: Let the initial kinetic energy be $KE_1 = KE$ and the final kinetic energy be $KE_2 = \frac{KE}{2}$ .

Step 4: Let the initial de Broglie wavelength be $\lambda_1$ and the final de Broglie wavelength be $\lambda_2$ .

Step 5: Write the ratio of the final to initial de Broglie wavelengths: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{KE}{\frac{KE}{2}}} = \sqrt{2}$

Step 6: Therefore, $\lambda_2 = \sqrt{2} \lambda_1$ . The de Broglie wavelength becomes $\sqrt{2}$ times.

Why other options are wrong:
- Option A (2 times): This is incorrect because the wavelength is inversely proportional to the square root of the kinetic energy, not directly proportional.
- **Option B ( $\frac{1}{{\sqrt 2 }}$ times): This is incorrect because the wavelength increases when kinetic energy decreases.
- Option C (4 times): This is incorrect because the relationship is with the square root of the kinetic energy, not the kinetic energy itself.

Key Concepts:
- de Broglie wavelength
- Kinetic energy
- Inverse proportionality

Memory Aids:**
- Remember the de Broglie wavelength formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$

Quick Tips:
- When dealing with changes in kinetic energy and wavelength, remember the inverse square root relationship.

189 Easy

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Application +4 | -1

The energy of an electron in the first Bohr orbit of H atom is -13.6 eV. The possible energy value of the first excited state for electron in Bohr orbit of hydrogen is:

View answer and explanation

Correct option: A

– 3.4 eV

Option A (– 3.4 eV) is correct because it represents the energy of the electron in the first excited state (n=2) of the hydrogen atom.

Solution approach

Correct Answer + Core Reason: – 3.4 eV. The energy of the first excited state is calculated using the Bohr model.

Step-by-Step Solution:

Step 1: Recall the energy formula for an electron in the nth Bohr orbit of a hydrogen atom: $E_n = \frac{-13.6}{n^2} eV$

Step 2: Identify the first excited state: n = 2

Step 3: Substitute n = 2 into the formula: $E_2 = \frac{-13.6}{2^2} eV = \frac{-13.6}{4} eV$

Step 4: Calculate the energy: $E_2 = -3.4 eV$

Step 5: Final Answer: -3.4 eV

Why other options are wrong:
- Option B: Incorrect energy value. Likely a miscalculation or confusion with a different energy level.
- Option C: Incorrect energy value. Likely a miscalculation or confusion with a different energy level.
- Option D: Incorrect sign and energy value. The energy should be negative, and the value is incorrect.

Key Concepts:
- Bohr model
- Energy levels of hydrogen atom

Memory Aids:
- Remember the energy formula: $E_n = \frac{-13.6}{n^2} eV$

Quick Tips:
- The first excited state corresponds to n=2.

190 Easy

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Application +4 | -1

The momentum of a photon of wavelength 6626

\mathrm{nm}

will be

View answer and explanation

Correct option: A

10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}

Option A (

10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}

) is correct because the momentum of a photon is calculated using the formula p = h/λ, where h is Planck's constant and λ is the wavelength.

Solution approach

Correct Answer + Core Reason: $10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$ . The momentum of a photon is calculated using $p=\frac{h}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given values: $\lambda = 6626 \mathrm{nm} = 6626 \times 10^{-9} \mathrm{m}$ , $h = 6.626 \times 10^{-34} \mathrm{Js}$

Step 2: Use the formula for momentum: $p = \frac{h}{\lambda}$

Step 3: Substitute the values: $p = \frac{{6.626 \times 10^{-34}}}{{6626 \times 10^{-9}}}$

Step 4: Simplify: $p = \frac{{6.626}}{{6626}} \times \frac{{10^{-34}}}{{10^{-9}}} = 10^{-34+9} = 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$

Step 5: Final Answer: $10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$

Why other options are wrong:
- **Option B ( $10^{-25} \mathrm{~kg} \mathrm{~ms}^{-1}$ ): This is incorrect due to a calculation error in the exponent.
- Option C ( $10^{-31} \mathrm{~kg} \mathrm{~m}^{-1}$ ): This is incorrect due to a calculation error in the exponent and incorrect units.
- Option D (zero): This is incorrect because a photon with a defined wavelength has momentum.

Key Concepts:
- Momentum of a photon
- Planck's constant
- Wavelength

Memory Aids:**
- $p = \frac{h}{\lambda}$

Quick Tips:
- Remember the formula for momentum of a photon.
- Pay attention to units and convert them if necessary.

191 Easy

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Analyzing +4 | -1

The electronic configuration of an element is

1{s^2},2{s^2}\,2{p^6},3{s^2}\,3{p^6}\,3{d^5},4{s^1}.\,

This represent its:

View answer and explanation

Correct option: D

Both 2 and 3

Option D is correct because the given electronic configuration represents the ground state of Chromium (Cr) and can also represent the cationic form of Manganese (Mn+).

Solution approach

Correct Answer + Core Reason: Both 2 and 3. The electronic configuration represents the ground state of Chromium and the cationic form of Manganese.

Step-by-Step Solution:

Step 1: Analyze the given electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$

Step 2: Determine the total number of electrons: 2 + 2 + 6 + 2 + 6 + 5 + 1 = 24

Step 3: Identify the element: The element with 24 electrons is Chromium (Cr).

Step 4: Check if the configuration is the ground state: The expected configuration for Cr is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^4 4s^2$ . However, due to the stability of half-filled d-orbitals, one electron from the 4s orbital is promoted to the 3d orbital, resulting in the configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ . This is the ground state configuration of Cr.

Step 5: Consider the cationic form: If the element has 25 electrons (Manganese, Mn), and it loses one electron to form Mn+, it will have 24 electrons. The electronic configuration of Mn is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$ . When it loses one electron, it becomes Mn+ with the configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ .

Step 6: Conclusion: The given electronic configuration represents the ground state of Cr and the cationic form of Mn+.

Why other options are wrong:
- Option A: Incorrect because it can also represent the cationic form.
- Option B: Incorrect because it can also represent the cationic form.
- Option C: Incorrect because it can also represent the ground state.

Key Concepts:
- Electronic configuration
- Ground state
- Cationic form
- Stability of half-filled orbitals

Memory Aids:
- Remember the exceptions to Hund's rule for Cr and Cu.

Quick Tips:
- Determine the number of electrons from the electronic configuration.
- Identify the element based on the number of electrons.
- Check for exceptions to Hund's rule.
- Consider the possibility of cationic or anionic forms.

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Remembering +4 | -1

The number of electrons accommodated in an orbit with principle quantum number 2, is

View answer and explanation

Correct option: D

8

Option D (8) is correct because the number of electrons accommodated in an orbit is given by

2n^2

, where n is the principal quantum number. For n=2, the number of electrons is

2(2^2) = 8

.

Solution approach

Correct Answer + Core Reason: 8. The formula $2n^2$ gives the maximum number of electrons.

Step-by-Step Solution:

Step 1: State the formula for the maximum number of electrons in an orbit: The maximum number of electrons in an orbit with principal quantum number n is given by $2n^2$ .

Step 2: Substitute n = 2 into the formula: $2(2^2) = 2(4) = 8$

Step 3: Conclude that the number of electrons accommodated in an orbit with n = 2 is 8.

Why other options are wrong:
- Option A: 2 is the number of electrons in the first orbit (n=1).
- Option B: 6 is the number of electrons in p-orbitals of the second shell, but the question asks for the total number of electrons in the orbit.
- Option C: 10 is the number of electrons in d-orbitals, which starts from the third shell.
- Option D: 8 is the correct number of electrons in the second orbit (n=2).

Key Concepts:
- Principal quantum number
- Electronic configuration
- Atomic structure

Memory Aids:
- $2n^2$ = Max electrons

Quick Tips:
- Remember the formula for calculating the maximum number of electrons in an orbit.

193 Easy

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Application +4 | -1

Which has the maximum value of magnetic moment?

View answer and explanation

Correct option: D

\mathrm{Fe}^{3+}

Option D (

\mathrm{Fe}^{3+}

) is correct because it has the maximum number of unpaired electrons, leading to the highest magnetic moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}^{3+}$ . It has the maximum number of unpaired electrons, leading to the highest magnetic moment.

Step-by-Step Solution:

Step 1: Write the electronic configurations of each ion:
$\mathrm{Mg}^{2+}: [Ne] = 1s^2 2s^2 2p^6$ (0 unpaired electrons)
$\mathrm{Ti}^{3+}: [Ar] 3d^1$ (1 unpaired electron)
$\mathrm{V}^{3+}: [Ar] 3d^2$ (2 unpaired electrons)
$\mathrm{Fe}^{3+}: [Ar]$

194 Easy

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Understanding +4 | -1

An isostere is

View answer and explanation

Correct option: A

{NO}_2^{-}

and

{O}_3

Option A (

{NO}_2^{-}

and

{O}_3

) is correct because they have the same number of atoms and the same number of electrons.

Solution approach

Correct Answer + Core Reason: ${NO}_2^{-}$ and ${O}_3$ . Isosteres have the same number of atoms and electrons.

Step-by-Step Solution:

Step 1: Define isosteres. Isosteres are molecules or ions that have the same number of atoms and the same number of electrons.

Step 2: Analyze option A: ${NO}_2^{-}$ and ${O}_3$ .
- ${NO}_2^{-}$ has 1 N atom + 2 O atoms = 3 atoms.
- ${O}_3$ has 3 O atoms = 3 atoms.
- ${NO}_2^{-}$ has 7 (N) + 2 * 8 (O) + 1 (charge) = 7 + 16 + 1 = 24 electrons.
- ${O}_3$ has 3 * 8 (O) = 24 electrons.

Step 3: Analyze other options (for completeness, though not strictly necessary for the solution).

Step 4: Conclude that ${NO}_2^{-}$ and ${O}_3$ are isosteres.

Why other options are wrong:
- **Option B ( ${NO}_2^{-}$ and ${PO}_4^{3-}$ ):** ${NO}_2^{-}$ has 3 atoms and 24 electrons. ${PO}_4^{3-}$ has 5 atoms and 50 electrons. They are not isosteres.
- **Option C ( ${CO}_2, {~N}_2 {O}, {NO}_3^{-}$ ):** ${CO}_2$ has 3 atoms and 22 electrons. ${N}_2{O}$ has 3 atoms and 22 electrons. ${NO}_3^{-}$ has 4 atoms and 24 electrons. While CO2 and N2O are isosteres, NO3- is not isosteric with them.
- **Option D ( ${ClO}_4^{-}$ and ${OCN}^{-}$ ):** ${ClO}_4^{-}$ has 5 atoms and 42 electrons. ${OCN}^{-}$ has 3 atoms and 16 electrons. They are not isosteres.

Key Concepts:
- Isosteres
- Counting atoms and electrons in molecules and ions

Memory Aids:
- Remember the definition of isosteres: same number of atoms and electrons.

Quick Tips:
- When identifying isosteres, carefully count the number of atoms and electrons in each species.

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Remembering +4 | -1

Zeeman effect explains splitting of lines in

View answer and explanation

Correct option: A

Magnetic Field

Option A (Magnetic Field) is correct because the Zeeman effect describes the splitting of spectral lines when an atom is placed in a magnetic field.

Solution approach

Correct Answer + Core Reason: Magnetic Field. Zeeman effect explains splitting of lines in magnetic field.

Step-by-Step Solution:
Step 1: Recall the definition of the Zeeman effect.
Step 2: The Zeeman effect is the splitting of spectral lines due to the presence of a magnetic field.

Why other options are wrong:
- Option B: The Stark effect describes the splitting of spectral lines in an electric field.
- Option C: The Zeeman effect is specific to magnetic fields, not both electric and magnetic fields.
- Option D: The Zeeman effect does explain the splitting of lines in a magnetic field.

Key Concepts:
- Zeeman effect
- Magnetic field
- Spectral lines

Memory Aids:
- Zeeman = Magnetic

Quick Tips:
- Remember the difference between the Zeeman and Stark effects.

196 Easy

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Remembering +4 | -1

Degeneracy of s, p, d, f and g sub-orbits respectively are

View answer and explanation

Correct option: B

1,3, 5, 7, 9

Option B (1,3, 5, 7, 9) is correct because the degeneracy of s, p, d, f, and g sub-orbits corresponds to the number of orbitals, which is given by 2l+1.

Solution approach

Correct Answer + Core Reason: 1,3, 5, 7, 9. The degeneracy is determined by the number of orbitals, which is 2l+1.

Step-by-Step Solution:

Step 1: Identify the azimuthal quantum number (l) for each sub-orbit.
- s: l = 0
- p: l = 1
- d: l = 2
- f: l = 3
- g: l = 4

Step 2: Calculate the degeneracy (number of orbitals) using the formula: Degeneracy = 2l + 1

Step 3: Apply the formula for each sub-orbit:
- s: 2(0) + 1 = 1
- p: 2(1) + 1 = 3
- d: 2(2) + 1 = 5
- f: 2(3) + 1 = 7
- g: 2(4) + 1 = 9

Step 4: The degeneracies are 1, 3, 5, 7, and 9 for s, p, d, f, and g sub-orbits, respectively.

Why other options are wrong:
- Option A: This option incorrectly assigns the degeneracy values. It does not follow the 2l+1 rule.
- Option C: This option suggests that all sub-orbits have the same degeneracy, which is incorrect.
- Option D: This option incorrectly assigns the degeneracy value for the g sub-orbit. It does not follow the 2l+1 rule.

Key Concepts:
- Degeneracy of orbitals
- Azimuthal quantum number (l)
- Number of orbitals (2l+1)

Memory Aids:
- Remember the l values for s, p, d, f, g: 0, 1, 2, 3, 4. Then use 2l+1 to find the degeneracy.

Quick Tips:
- Always remember that the degeneracy is equal to the number of orbitals, which is given by the formula 2l+1.

197 Easy

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Understanding +4 | -1

The de-Broglie equation applies

View answer and explanation

Correct option: C

All the material objects in motion

Option C (All the material objects in motion) is correct because the de Broglie equation applies to all matter, relating wavelength to momentum.

Solution approach

Correct Answer + Core Reason: All the material objects in motion. De Broglie equation applies to all matter.

Step-by-Step Solution:

Step 1: State the de Broglie equation: $\lambda = \frac{h}{p} = \frac{h}{mv}$ , where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $p$ is momentum, $m$ is mass, and $v$ is velocity.

Step 2: Explain the significance of the de Broglie equation. The equation demonstrates the wave-particle duality of matter, meaning that all matter exhibits both wave-like and particle-like properties.

Step 3: Determine the applicability of the de Broglie equation. The equation applies to all material objects in motion, regardless of their size or composition.

Why other options are wrong:
- Option A (To protons only): The de Broglie equation applies to all matter, not just protons.
- Option B (To electrons only): The de Broglie equation applies to all matter, not just electrons.
- Option D (To neutrons only): The de Broglie equation applies to all matter, not just neutrons.

Key Concepts:
- De Broglie equation
- Wave-particle duality
- Momentum

Memory Aids:
- Remember the de Broglie equation: $\lambda = \frac{h}{mv}$

Quick Tips:
- The de Broglie equation is a fundamental concept in quantum mechanics.

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Application +4 | -1

The energy of an electron in Bohr's orbit of hydrogen atom is

-13.6 \mathrm{~eV}

. The total electronic energy of a hypothetical

\mathrm{He}

atom in which there are no electron-electron repulsions or interactions is

View answer and explanation

Correct option: C

-108.8 \mathrm{~eV}

Option C (

-108.8 \mathrm{~eV}

) is correct because the total electronic energy is calculated by multiplying the energy of one electron by the square of the atomic number and the number of electrons.

Solution approach

Correct Answer + Core Reason: $-108.8 \mathrm{~eV}$ . The total electronic energy is calculated considering the energy of each electron and the atomic number.

Step-by-Step Solution:

Step 1: The energy of an electron in the Bohr's orbit of a hydrogen atom is given as $-13.6 \mathrm{~eV}$ .

Step 2: For a hypothetical He atom with no electron-electron repulsions, each electron experiences the full nuclear charge of +2.

Step 3: The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 \frac{{Z^2}}{{n^2}} \mathrm{~eV}$ , where $Z$ is the atomic number and $n$ is the principal quantum number.

Step 4: For He, $Z = 2$ . Since we are considering the ground state, $n = 1$ .\nSo, the energy of one electron in the hypothetical He atom is $E_1 = -13.6 \frac{{2^2}}{{1^2}} = -13.6 \times 4 = -54.4 \mathrm{~eV}$ .

Step 5: Since there are two electrons in the He atom, the total electronic energy is $2 \times (-54.4 \mathrm{~eV}) = -108.8 \mathrm{~eV}$ .

Why other options are wrong:
- Option A: $27.2 \mathrm{~eV}$ : This is incorrect because it doesn't account for the square of the atomic number and the negative sign for energy.
- Option B: $-27.2 \mathrm{~eV}$ : This is incorrect because it only multiplies the hydrogen energy by 2 and doesn't account for the increased nuclear charge.
- Option D: $108 \mathrm{~eV}$ : This is incorrect because it has the wrong sign and a slightly different magnitude due to rounding errors.

Key Concepts:
- Bohr's Model of the Atom
- Energy Levels of Hydrogen-like Species
- Atomic Number

Memory Aids:
- Energy of an electron in nth orbit: $E_n = -13.6 \frac{{Z^2}}{{n^2}} \mathrm{~eV}$

Quick Tips:
- Remember to account for the atomic number when calculating the energy of electrons in atoms other than hydrogen.
- Ensure you multiply by the number of electrons to get the total electronic energy.

199 Easy

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Understanding +4 | -1

Which wave property is directly proportional to energy of electromagnetic radiation:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because energy of electromagnetic radiation is directly proportional to frequency, wave number and inversely proportional to wavelength. Velocity is constant.

Solution approach

Correct Answer + Core Reason: All of these. Energy is directly proportional to frequency and wave number.

Step-by-Step Solution:

Step 1: Recall the relationship between energy (E), frequency (ν), wavelength (λ), and wave number (\bar{ν}) of electromagnetic radiation:
- $E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}$

Step 2: Analyze the relationships:
- Energy (E) is directly proportional to frequency (ν): $E \propto \nu$
- Energy (E) is inversely proportional to wavelength (λ): $E \propto \frac{1}{\lambda}$
- Energy (E) is directly proportional to wave number (\bar{ν}): $E \propto \bar{\nu}$

Step 3: Velocity (c) is constant for electromagnetic radiation in a vacuum.

Step 4: Therefore, all of the given wave properties (frequency, wave number) are directly proportional to the energy of electromagnetic radiation.

Why other options are wrong:
- Option A (Velocity): Velocity is constant and not directly proportional to energy.
- Option B (Frequency): Frequency is directly proportional to energy, but this is not the only correct answer.
- Option C (Wave number): Wave number is directly proportional to energy, but this is not the only correct answer.

Key Concepts:
- Electromagnetic radiation
- Energy, frequency, wavelength, and wave number relationship

Memory Aids:
- Remember the formula: $E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}$

Quick Tips:
- Understand the relationships between the different properties of electromagnetic radiation.

200 Easy

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Understanding +4 | -1

The ratio of specific charge of a neutron and an

\alpha

-particle is

View answer and explanation

Correct option: B

0

Option B (0) is correct because a neutron has no charge, so its specific charge (charge/mass) is zero, making the ratio zero as well.

Solution approach

Correct Answer + Core Reason: 0. Neutron has no charge.

Step-by-Step Solution:

Step 1: Define specific charge. Specific charge is the ratio of charge to mass (q/m).

Step 2: Determine the charge of a neutron. A neutron has a charge of 0.

Step 3: Determine the charge of an alpha particle. An alpha particle has a charge of +2.

Step 4: Calculate the specific charge of a neutron. Since the charge is 0, the specific charge is 0/m = 0.

Step 5: Calculate the specific charge of an alpha particle. The specific charge is 2/m, where m is the mass of the alpha particle.

Step 6: Determine the ratio of the specific charge of a neutron to an alpha particle. The ratio is 0 / (2/m) = 0.

Why other options are wrong:
- Option A: $2:1$ is incorrect because the neutron has no charge.
- Option C: $1:4$ is incorrect because the neutron has no charge.
- Option D: $1:1$ is incorrect because the neutron has no charge.

Key Concepts:
- Specific charge
- Neutron charge
- Alpha particle charge

Memory Aids:
- Neutron = neutral = no charge

Quick Tips:
- Remember that neutrons have no charge.

201 Easy

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Remembering +4 | -1

When an electron moves from higher orbit to a lower orbit … is produced

View answer and explanation

Correct option: B

Emission spectra

Option B (Emission spectra) is correct because when an electron transitions from a higher energy orbit to a lower energy orbit, it emits energy in the form of photons, resulting in an emission spectrum.

Solution approach

Correct Answer + Core Reason: Emission spectra. The difference of energy is given out.

Step-by-Step Solution:
Step 1: Recall the Bohr model of the atom.
Step 2: When an electron transitions from a higher energy level to a lower energy level, it releases energy.
Step 3: This energy is emitted as a photon of light.
Step 4: The collection of these emitted photons at different wavelengths forms an emission spectrum.

Why other options are wrong:
- Option A (Absorption spectra): Absorption spectra occur when an electron absorbs energy to move to a higher orbit.
- **Option C ( $\alpha -$ particle): Alpha particles are helium nuclei and are not produced during electronic transitions.
- Option D (None of these): Emission spectra is the correct answer.

Key Concepts:
- Bohr model of the atom
- Electronic transitions
- Emission spectra
- Absorption spectra

Memory Aids:
- Emission: electron emits energy and goes to lower orbit.
- Absorption: electron absorbs energy and goes to higher orbit.

Quick Tips:**
- Understand the difference between emission and absorption spectra.

202 Easy

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Remembering +4 | -1

Which one of the following orbitals is spherical in shape

View answer and explanation

Correct option: A

4s

Option A (4s) is correct because s-orbitals are always spherical in shape, regardless of the principal quantum number.

Solution approach

Correct Answer + Core Reason: 4s. S-orbitals are spherically symmetrical.

Step-by-Step Solution:

Step 1: Recall the shapes of atomic orbitals.

Step 2: s-orbitals are spherical.\np-orbitals are dumbbell-shaped.\nd-orbitals have more complex shapes (e.g., cloverleaf).\nf-orbitals have even more complex shapes.

Step 3: Identify the orbital with a spherical shape.

Step 4: The 4s orbital is spherical.

Why other options are wrong:
- Option B: 3p orbitals are dumbbell-shaped, not spherical.
- Option C: 3d orbitals have more complex shapes, such as a cloverleaf shape, and are not spherical.
- Option D: 4f orbitals have very complex shapes and are not spherical.

Key Concepts:
- Atomic orbitals
- Shapes of orbitals (s, p, d, f)

Memory Aids:
- s = spherical

Quick Tips:
- Remember the basic shapes of s, p, and d orbitals.

203 Easy

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Remembering +4 | -1

Angular momentum of an electron in an orbital is given by

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the angular momentum of an electron in an orbital is given by the formula

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

.

Solution approach

Correct Answer + Core Reason: $\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}$ . This is the formula for angular momentum of an electron in an orbital.

Step-by-Step Solution:

Step 1: Recall the formula for the angular momentum of an electron in an orbital.

Step 2: The angular momentum of an electron in an orbital is given by the formula: $\sqrt{l(l+1)} \frac{h}{2\pi}$ , where $l$ is the azimuthal quantum number and $h$ is Planck's constant.

Step 3: Compare the formula with the given options.

Why other options are wrong:
- Option A: $n\frac{h}{{2\pi }}$ represents the angular momentum in Bohr's model, not the angular momentum of an electron in an orbital.
- Option C: $n\frac{h}{{4\pi }}$ is not a correct formula for angular momentum.
- Option D: The correct formula is given in option B.

Key Concepts:
- Angular momentum of an electron in an orbital
- Azimuthal quantum number (l)
- Planck's constant (h)

Memory Aids:
- Angular momentum = $\sqrt{l(l+1)} \frac{h}{2\pi}$ .

Quick Tips:
- Know the formulas for different types of angular momentum.

204 Easy

204 of 239

Application +4 | -1

What are the total numbers of orbitals and electrons for

m = 0

, in the ground state of

Zn

(atomic number = 30).

View answer and explanation

Correct option: A

7 orbitals, 14 electrons

Option A (7 orbitals, 14 electrons) is correct because in the ground state of Zinc, there are 7 orbitals with m=0, each containing 2 electrons, totaling 14 electrons.

Solution approach

Correct Answer + Core Reason: 7 orbitals, 14 electrons. The number of orbitals and electrons with m=0 is determined by the electronic configuration of Zinc.

Step-by-Step Solution:

Step 1: Write the electronic configuration of Zinc (Zn): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10}$ .

Step 2: Identify the orbitals with $m = 0$ :
- $1s$ : 1 orbital with $m = 0$
- $2s$ : 1 orbital with $m = 0$
- $2p$ : 1 orbital with $m = 0$ (the $p_z$ orbital)
- $3s$ : 1 orbital with $m = 0$
- $3p$ : 1 orbital with $m = 0$ (the $p_z$ orbital)
- $4s$ : 1 orbital with $m = 0$
- $3d$ : 1 orbital with $m = 0$ (the $d_{z^2}$ orbital)

Step 3: Count the total number of orbitals with $m = 0$ : $1 + 1 + 1 + 1 + 1 + 1 + 1 = 7$ orbitals.

Step 4: Determine the number of electrons in each orbital: Each orbital can hold 2 electrons.

Step 5: Calculate the total number of electrons with $m = 0$ : $7 \times 2 = 14$ electrons.

Final Answer: 7 orbitals, 14 electrons

Why other options are wrong:
- Option B (6 orbitals, 12 electrons): This is incorrect because it misses one of the orbitals with m=0.
- Option C (5 orbitals, 10 electrons): This is incorrect because it misses two of the orbitals with m=0.
- Option D (3 orbitals, 6 electrons): This is incorrect because it significantly underestimates the number of orbitals with m=0.

Key Concepts:
- Electronic Configuration
- Atomic Orbitals
- Magnetic Quantum Number (m)
- Hund's Rule

Memory Aids:
- Remember the shapes and orientations of s, p, and d orbitals.
- The magnetic quantum number (m) describes the orientation of the orbital in space.

Quick Tips:
- Always write the complete electronic configuration first.
- Identify the orbitals with m=0 carefully.

205 Easy

205 of 239

Application +4 | -1

The electronic configurations of four elements L,P,Q and R are given below,

L = 1{s^2},2{s^2}2{p^4}\;\;\;Q = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^5}

,

P = 1{s^2},2{s^2}2{p^6},3{s^1}\;\;\;R = 1{s^2},2{s^2}2{p^6},3{s^2}

, The formula of the ionic compounds that can be formed between these elements are:

View answer and explanation

Correct option: C

{P_2}L,\;RL,\;PQ,\;R{Q_2}

Option C (

{P_2}L,\;RL,\;PQ,\;R{Q_2}

) is correct because it represents the correct ionic formulas based on the valencies of the elements L, P, Q, and R.

Solution approach

Correct Answer + Core Reason: ${P_2}L,\;RL,\;PQ,\;R{Q_2}$ . The formulas are determined by the valencies of the elements.

Step-by-Step Solution:

Step 1: Determine the valencies of the elements based on their electronic configurations:
- L: $1s^2 2s^2 2p^4$ (Oxygen family, needs 2 electrons to complete octet, valency = 2-)
- P: $1s^2 2s^2 2p^6 3s^1$ (Alkali metal, loses 1 electron, valency = 1+)
- Q: $1s^2 2s^2 2p^6 3s^2 3p^5$ (Halogen, needs 1 electron to complete octet, valency = 1-)
- R: $1s^2 2s^2 2p^6 3s^2$ (Alkaline earth metal, loses 2 electrons, valency = 2+)

Step 2: Form the ionic compounds:
- L and P: $P^+ L^{2-}$ -> ${P_2}L$
- R and L: $R^{2+} L^{2-}$ -> $RL$
- P and Q: $P^+ Q^-$ -> $PQ$
- R and Q: $R^{2+} Q^-$ -> $R{Q_2}$

Step 3: Final Answer: ${P_2}L,\;RL,\;PQ,\;R{Q_2}$

Why other options are wrong:
- **Option A ( ${L_2}P,\;RL,\;PQ,\;{R_2}Q$ ):** Incorrect because the formula between L and P should be ${P_2}L$ , and the formula between R and Q should be $R{Q_2}$ .
- **Option B ( $LP,\;RL,\;PQ,\;RQ$ ):** Incorrect because the formula between L and P should be ${P_2}L$ , and the formula between R and Q should be $R{Q_2}$ .
- **Option D ( $LP,\;{R_2}L,\;{P_2}Q,\;RQ$ ):** Incorrect because the formula between L and P should be ${P_2}L$ , the formula between R and L should be $RL$ , the formula between P and Q should be $PQ$ , and the formula between R and Q should be $R{Q_2}$ .

Key Concepts:
- Electronic configuration
- Valency
- Ionic compounds
- Formation of ionic compounds

Memory Aids:
- Remember the octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons.
- Cations are positive ions, anions are negative ions.

Quick Tips:
- Determine the valency of each element first.
- Balance the charges in the ionic compound.

206 Easy

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Application +4 | -1

The energies of the electron in a hydrogen like ion are given by :

E_n=-\left(2.18 \times 10^{-18}\right) Z^2\left(\frac{1}{n^2}\right)

, where

n

is principal quantum number and

Z

is the atomic number of the element. The ionisation energy (in

\mathrm{kJ} \mathrm{mol}^{-1}

) of the

\mathrm{Li}^{2+}

ion is:

View answer and explanation

Correct option: D

11.8 \times 10^3

The correct answer is

11.8 \times 10^3

because it is calculated using the given formula for ionization energy with Z=3 and n=1, then converting to kJ/mol.

Solution approach

Correct Answer + Core Reason: $11.8 \times 10^3$ . Ionization energy is calculated using the given formula with Z=3 and n=1, then converted to kJ/mol.

Step-by-Step Solution:

Step 1: Identify the given formula: $E_n=-\left(2.18 \times 10^{-18}\right) Z^2\left(\frac{{1}}{{n^2}}\right)$ J/atom

Step 2: For Li $^{2+}$ , Z = 3 and n = 1 (ionization from the ground state).

Step 3: Calculate the ionization energy per atom: $E = -\left(2.18 \times 10^{-18}\right) (3^2) \left(\frac{{1}}{{1^2}}\right) = -\left(2.18 \times 10^{-18}\right) (9) = -19.62 \times 10^{-18}$ J/atom. Since we are looking for ionization energy, we take the absolute value: $19.62 \times 10^{-18}$ J/atom.

Step 4: Convert to J/mol: $19.62 \times 10^{-18} \text{{ J/atom }} \times 6.022 \times 10^{23} \text{{ atoms/mol }} = 11.81 \times 10^6 \text{{ J/mol}}$

Step 5: Convert to kJ/mol: $11.81 \times 10^6 \text{{ J/mol }} \times \frac{{1 \text{{ kJ}}}}{{1000 \text{{ J}}}} = 11.81 \times 10^3 \text{{ kJ/mol}}$

Step 6: Final Answer: $11.8 \times 10^3$ kJ/mol

Why other options are wrong:
- **Option A ( $1.18 \times 10^3$ ): This is incorrect because it doesn't account for the Z² term correctly. It might be a calculation error.
- Option B ( $118 \times 10^3$ ): This is incorrect due to a miscalculation, likely not converting from J/mol to kJ/mol correctly or an error in the exponent.
- Option C ( $0.118 \times 10^3$ ): This is incorrect because it significantly underestimates the ionization energy. It's likely a result of dividing instead of multiplying by Avogadro's number and not squaring Z.

Key Concepts:
- Ionization Energy
- Bohr's Model
- Atomic Number

Memory Aids:
- Remember the formula for energy levels in hydrogen-like ions.
- Pay attention to units (J vs kJ).

Quick Tips:**
- Always double-check your calculations, especially when dealing with exponents.
- Remember to convert units to match the desired output.

207 Easy

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Remembering +4 | -1

The shape of

p

-orbital is

View answer and explanation

Correct option: C

Dumb-bell

Option C (Dumb-bell) is correct because the p-orbital has a dumbbell shape.

Solution approach

Correct Answer + Core Reason: Dumb-bell. The shape of a p-orbital is dumbbell-shaped.

Step-by-Step Solution:

Step 1: Recall the shapes of atomic orbitals.

Step 2: The s-orbital is spherical, the p-orbital is dumbbell-shaped, and the d-orbital has more complex shapes.

Step 3: Identify the shape of the p-orbital as dumbbell.

Why other options are wrong:
- Option A: Elliptical is not the correct shape for any standard atomic orbital.
- Option B: Spherical is the shape of an s-orbital, not a p-orbital.
- Option D: Complex geometrical is too general and doesn't accurately describe the specific shape of a p-orbital. While d-orbitals have more complex shapes, p-orbitals are specifically dumbbell-shaped.

Key Concepts:
- Atomic orbitals (s, p, d, f)
- Shapes of atomic orbitals

Memory Aids:
- s = spherical, p = peanut/dumbbell

Quick Tips:
- Visualize the shapes of the orbitals to quickly answer this type of question.

208 Easy

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Understanding +4 | -1

STATEMENT-1 :

\gamma

-rays have very high penetrating power.

STATEMENT-2 :

\gamma

-rays are electromagnetic radiations of high energy.

View answer and explanation

Correct option: A

If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1

Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because gamma rays have high penetrating power due to their high energy electromagnetic radiation nature.

Solution approach

Correct Answer + Core Reason: If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1. Gamma rays have high penetrating power due to their high energy electromagnetic radiation nature.

Step-by-Step Solution:

Step 1: Analyze Statement 1: $\gamma$ -rays have very high penetrating power. This statement is true. Gamma rays are highly energetic and can penetrate through many materials.

Step 2: Analyze Statement 2: $\gamma$ -rays are electromagnetic radiations of high energy. This statement is also true. Gamma rays are a form of electromagnetic radiation with very high energy and short wavelengths.

Step 3: Determine if Statement 2 explains Statement 1. The high energy of gamma rays (Statement 2) is the reason why they have high penetrating power (Statement 1). Therefore, Statement 2 is the correct explanation of Statement 1.

Why other options are wrong:
- Option B: If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1: This is incorrect because Statement 2 directly explains Statement 1.
- Option C: If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE: This is incorrect because both statements are true.
- Option D: If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE: This is incorrect because both statements are true.

Key Concepts:
- Gamma Rays
- Electromagnetic Radiation
- Penetrating Power
- Energy of Radiation

Memory Aids:
- High energy radiation has high penetrating power.

Quick Tips:
- Remember that the energy of electromagnetic radiation is inversely proportional to its wavelength. Shorter wavelengths (like gamma rays) have higher energy and penetrating power.

209 Easy

209 of 239

Application +4 | -1

The radius of first Bohr's orbit for hydrogen is

0.53 \mathop {\text{A}}\limits^ \circ

. The radius of third Bohr's orbit would be

View answer and explanation

Correct option: D

4.77\mathop {\text{A}}\limits^ \circ

Option D (

4.77\mathop {\text{A}}\limits^ \circ

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

n^2

, so the third orbit's radius is 9 times the first.

Solution approach

Correct Answer + Core Reason: $4.77\mathop {\text{A}}\limits^ \circ$ . The radius of the nth Bohr orbit is proportional to $n^2$ , so the third orbit's radius is 9 times the first.

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n = r_1 \times n^2$ , where $r_1$ is the radius of the first Bohr orbit and n is the principal quantum number.

Step 2: Given: $r_1 = 0.53 \mathop {\text{A}}\limits^ \circ$ and n = 3 (third Bohr orbit)

Step 3: Substitute the values into the formula: $r_3 = 0.53 \mathop {\text{A}}\limits^ \circ \times 3^2 = 0.53 \mathop {\text{A}}\limits^ \circ \times 9$

Step 4: Calculate the radius of the third Bohr orbit: $r_3 = 4.77 \mathop {\text{A}}\limits^ \circ$

Step 5: Final Answer: $4.77\mathop {\text{A}}\limits^ \circ$

Why other options are wrong:
- **Option A ( $0.79 \mathop {\text{A}}\limits^ \circ$ ): This is incorrect because it likely involves an error in the calculation or an incorrect application of the formula.
- Option B ( $1.59 \mathop {\text{A}}\limits^ \circ$ ): This is incorrect because it likely involves an error in the calculation or an incorrect application of the formula.
- Option C ( $3.18 \mathop {\text{A}}\limits^ \circ$ ): This is incorrect because it likely involves an error in the calculation or an incorrect application of the formula.

Key Concepts:
- Bohr's Model
- Radius of Bohr Orbit

Memory Aids:**
- Remember that the radius is proportional to $n^2$ .

Quick Tips:
- Always use the correct formula for the radius of the Bohr orbit.
- Double-check the calculations to avoid errors.

210 Easy

210 of 239

Remembering +4 | -1

The energy of a photon is inversely proportional to its

View answer and explanation

Correct option: A

Wave length

Option A (Wave length) is correct because the energy of a photon is inversely proportional to its wavelength, as described by the equation E = hc/λ.

Solution approach

Correct Answer + Core Reason: Wave length. The energy of a photon is inversely proportional to its wavelength, according to the equation $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Recall the relationship between energy, frequency, and wavelength of a photon.

Step 2: Write down the equation: $E = hf = \frac{{hc}}{\lambda}$ , where E is energy, h is Planck's constant, f is frequency, c is the speed of light, and $\lambda$ is wavelength.

Step 3: From the equation, it is clear that energy is inversely proportional to wavelength.

Why other options are wrong:
- Option B: Frequency is directly proportional to energy ( $E = hf$ ).
- Option C: Wave number is directly proportional to energy (wave number is the reciprocal of wavelength).
- Option D: Velocity is constant for photons (speed of light).

Key Concepts:
- Energy of a photon
- Planck's equation: $E = hf$
- Relationship between frequency and wavelength: $c = f\lambda$

Memory Aids:
- E = hc/λ: Energy is inversely proportional to wavelength.

Quick Tips:
- Remember the relationship between energy, frequency, and wavelength.

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