STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Easy, Page 5

121 Easy

Remembering +4 | -1

The velocity of a photon is:

View answer and explanation

Correct option: A

Independent of its wavelength

Option A (Independent of its wavelength) is correct because the speed of a photon in a vacuum is a constant, denoted by 'c', and is independent of its wavelength.

Solution approach

Correct Answer + Core Reason: Independent of its wavelength. The speed of a photon in vacuum is constant.

Step-by-Step Solution:

Step 1: Recall that a photon is a particle of light, which is an electromagnetic wave.

Step 2: The speed of light in a vacuum is a fundamental constant, denoted by 'c', and its value is approximately $3 \times 10^8$ m/s.

Step 3: The speed of a photon in a vacuum is independent of its wavelength or frequency.

Why other options are wrong:
- Option B (Depends on its wavelength): The speed of a photon in a vacuum does not depend on its wavelength. The relationship between speed, frequency, and wavelength is $c = \lambda f$ , but 'c' is constant in a vacuum.
- Option C (Depends on its source): The speed of a photon in a vacuum does not depend on its source. It is a fundamental constant.
- Option D (Equal to square of its amplitude): The speed of a photon is not related to the square of its amplitude.

Key Concepts:
- Speed of light in a vacuum
- Photon
- Electromagnetic wave

Memory Aids:
- Remember that the speed of light in a vacuum is a constant 'c'.

Quick Tips:
- The speed of light in a vacuum is always the same, regardless of the photon's energy or wavelength.

122 Easy

122 of 176Question permalink

Application +4 | -1

When electrons in N-shell of excited hydrogen atom return to ground state, the number of possible lines spectrum is:

View answer and explanation

Correct option: A

6

Option A (6) is correct because the number of spectral lines is calculated using the formula n(n-1)/2 where n is the orbit number.

Solution approach

Correct Answer + Core Reason: 6. The number of spectral lines is calculated using the formula n(n-1)/2.

Step-by-Step Solution:

Step 1: Identify the initial and final energy levels: The electron transitions from the N-shell (n = 4) to the ground state (n = 1).

Step 2: Determine the number of possible transitions: The electron can transition from n = 4 to n = 3, n = 4 to n = 2, n = 4 to n = 1, n = 3 to n = 2, n = 3 to n = 1, and n = 2 to n = 1.

Step 3: Use the formula for the number of spectral lines: The number of spectral lines is given by $\frac{{n(n - 1)}}{2}$ , where $n$ is the initial energy level. In this case, $n = 4$ , so the number of spectral lines is $\frac{{4(4 - 1)}}{2} = \frac{{4 \times 3}}{2} = 6$ .

Step 4: Final Answer: 6

Why other options are wrong:
- Option B: 4: Incorrect, this is not the correct number of spectral lines.
- Option C: 2: Incorrect, this is not the correct number of spectral lines.
- Option D: 3: Incorrect, this is not the correct number of spectral lines.

Key Concepts:
- Hydrogen spectrum
- Spectral lines
- Electronic transitions

Memory Aids:
- Number of spectral lines: $\frac{{n(n - 1)}}{2}$

Quick Tips:
- Remember the formula for calculating the number of spectral lines when an electron transitions from a higher energy level to a lower energy level.

123 Easy

123 of 176Question permalink

Understanding +4 | -1

B{e^{^{ + 2}}}

is isoelectronic with which of the following ions

View answer and explanation

Correct option: B

L{i^ + }

Option B (

L{i^ + }

) is correct because both

B{e^{^{ + 2}}}

and

L{i^ + }

have 2 electrons, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: $L{i^ + }$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $B{e^{^{ + 2}}}$ . Beryllium (Be) has 4 electrons. $B{e^{^{ + 2}}}$ has lost 2 electrons, so it has 4 - 2 = 2 electrons.

Step 2: Determine the number of electrons in each of the options:
- A. ${H^ + }$ : Hydrogen (H) has 1 electron. ${H^ + }$ has lost 1 electron, so it has 1 - 1 = 0 electrons.
- B. $L{i^ + }$ : Lithium (Li) has 3 electrons. $L{i^ + }$ has lost 1 electron, so it has 3 - 1 = 2 electrons.
- C. $N{a^ + }$ : Sodium (Na) has 11 electrons. $N{a^ + }$ has lost 1 electron, so it has 11 - 1 = 10 electrons.
- D. $M{g^{2 + }}$ : Magnesium (Mg) has 12 electrons. $M{g^{2 + }}$ has lost 2 electrons, so it has 12 - 2 = 10 electrons.

Step 3: Identify the ion with the same number of electrons as $B{e^{^{ + 2}}}$ . $L{i^ + }$ has 2 electrons.

Why other options are wrong:
- Option A: ${H^ + }$ has 0 electrons, not 2.
- Option C: $N{a^ + }$ has 10 electrons, not 2.
- Option D: $M{g^{2 + }}$ has 10 electrons, not 2.

Key Concepts:
- Isoelectronic species
- Number of electrons in ions

Memory Aids:
- Iso- means same. Isoelectronic means same number of electrons.

Quick Tips:
- Remember to account for the charge when determining the number of electrons in an ion.

124 Easy

124 of 176Question permalink

Application +4 | -1

Calculate the number of a and particles in the following change.

{ }_{92}^{235} \mathrm{U} \rightarrow \frac{{207}}{{{\text{a}}2}}{\text{Pb}}

View answer and explanation

Correct option: D

7

4

Option D (7, 4) is correct because the change in mass number is 235-207 = 28, and each alpha particle emission reduces the mass number by 4, so 28/4 = 7 alpha particles are emitted. The change in atomic number is 92-82 = 10, and each alpha particle reduces the atomic number by 2, and each beta particle increases the atomic number by 1. Thus, 10 = 7*2 - x*(-1) => x = 4.

Solution approach

Correct Answer + Core Reason: 7 4. Alpha decay reduces mass number by 4 and atomic number by 2, while beta decay increases atomic number by 1.

Step-by-Step Solution:

Step 1: The nuclear reaction is: ${}_{92}^{235} \mathrm{U} \rightarrow {}_{82}^{207} \mathrm{Pb} + x \alpha + y \beta$

Step 2: For alpha particles, the mass number changes by 4 and the atomic number changes by 2.\nFor beta particles, the mass number remains the same and the atomic number increases by 1.

Step 3: Change in mass number: 235 - 207 = 28. Each alpha particle has a mass number of 4, so $4x = 28$ . Therefore, $x = \frac{{28}}{4} = 7$ alpha particles.

Step 4: Change in atomic number: 92 - 82 = 10. Each alpha particle reduces the atomic number by 2, and each beta particle increases the atomic number by 1. So, $2x - y = 10$ , where x is the number of alpha particles and y is the number of beta particles.

Step 5: Substitute x = 7: $2(7) - y = 10$ => $14 - y = 10$ => $y = 14 - 10 = 4$ beta particles.

Step 6: Therefore, there are 7 alpha particles and 4 beta particles.

Why other options are wrong:
- Option A: Incorrect number of alpha and beta particles.
- Option B: Incorrect number of alpha and beta particles.
- Option C: Incorrect number of alpha and beta particles.

Key Concepts:
- Alpha decay
- Beta decay
- Nuclear reactions
- Balancing nuclear equations

Memory Aids:
- Alpha decay: ${}_2^4 \mathrm{He}$
- Beta decay: ${}_{-1}^0 \mathrm{e}$

Quick Tips:
- Balance the mass number and atomic number separately.

125 Easy

125 of 176Question permalink

Application +4 | -1

Which atom has as many as s-electrons as p-electrons?

View answer and explanation

Correct option: B

Mg

Option B (Mg) is correct because Magnesium (Mg) has the electronic configuration 1s² 2s² 2p⁶ 3s², which has 6 s-electrons and 6 p-electrons.

Solution approach

Correct Answer + Core Reason: Mg. Magnesium has an equal number of s and p electrons in its electronic configuration.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each atom.

Step 2: Count the number of s and p electrons in each configuration.

Step 3: H: 1s¹ (1 s-electron, 0 p-electrons)\nMg: 1s² 2s² 2p⁶ 3s² (6 s-electrons, 6 p-electrons)\nN: 1s² 2s² 2p³ (4 s-electrons, 3 p-electrons)\nNa: 1s² 2s² 2p⁶ 3s¹ (5 s-electrons, 6 p-electrons)

Step 4: Identify the atom with an equal number of s and p electrons.

Why other options are wrong:
- Option A: H is incorrect because it has 1 s-electron and 0 p-electrons.
- Option C: N is incorrect because it has 4 s-electrons and 3 p-electrons.
- Option D: Na is incorrect because it has 5 s-electrons and 6 p-electrons.

Key Concepts:
- Electronic configuration
- s and p electrons

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

Quick Tips:
- Always write the complete electronic configuration to accurately count the number of s and p electrons.

126 Easy

126 of 176Question permalink

Understanding +4 | -1

The ratio of charge and mass would be greter for

View answer and explanation

Correct option: B

Electron

Option B (Electron) is correct because electrons have a very small mass and a significant charge, resulting in a high charge-to-mass ratio compared to protons, neutrons, and alpha particles.

Solution approach

Correct Answer + Core Reason: Electron. Electron has the highest charge to mass ratio due to its very small mass.

Step-by-Step Solution:

Step 1: Recall the charges and masses of the given particles:
- Proton: Charge = +1, Mass ≈ 1 amu
- Electron: Charge = -1, Mass ≈ 1/1836 amu
- Neutron: Charge = 0, Mass ≈ 1 amu
- Alpha particle: Charge = +2, Mass ≈ 4 amu

Step 2: Calculate the charge-to-mass ratio for each particle (approximately):
- Proton: 1/1 ≈ 1
- Electron: 1/(1/1836) ≈ 1836
- Neutron: 0/1 = 0
- Alpha particle: 2/4 = 0.5

Step 3: Compare the charge-to-mass ratios. The electron has the highest ratio.

Why other options are wrong:
- Option A: Proton has a lower charge-to-mass ratio compared to an electron.
- Option C: Neutron has a charge of 0, so its charge-to-mass ratio is 0.
- Option D: Alpha particle has a lower charge-to-mass ratio compared to an electron.

Key Concepts:
- Charge-to-mass ratio
- Properties of subatomic particles

Memory Aids:
- Remember that electrons are much lighter than protons and neutrons.

Quick Tips:
- Understand the relative masses and charges of subatomic particles.

127 Easy

127 of 176Question permalink

Application +4 | -1

If the radius of first Bohr orbit is ‘a’, then de Broglie wavelength off electron in 3rd orbit is nearly

View answer and explanation

Correct option: B

6\pi a

Option B (

6\pi a

) is correct because the circumference of the 3rd Bohr orbit is

2\pi r = n\lambda

, and

r = n^2a

, so

2\pi(9a) = 3\lambda

, which gives

\lambda = 6\pi a

.

Solution approach

Correct Answer + Core Reason: $6\pi a$ . The circumference of the 3rd Bohr orbit is $2\pi r = n\lambda$ , and $r = n^2a$ , so $2\pi(9a) = 3\lambda$ , which gives $\lambda = 6\pi a$ .

Step-by-Step Solution:

Step 1: Recall Bohr's postulate that the circumference of an orbit is an integral multiple of the de Broglie wavelength:
$2\pi r = n\lambda$ , where:\nr is the radius of the orbit\nn is the principal quantum number (orbit number)
$\lambda$ is the de Broglie wavelength

Step 2: The radius of the nth Bohr orbit is given by:\nr = $n^2a$ , where 'a' is the radius of the first Bohr orbit.

Step 3: For the 3rd orbit, n = 3, so the radius is:\nr = $3^2a = 9a$

Step 4: Substitute n = 3 and r = 9a into the Bohr's postulate equation:
$2\pi (9a) = 3\lambda$

Step 5: Solve for the de Broglie wavelength $\lambda$ :
$\lambda = \frac{2\pi (9a)}{3} = 6\pi a$

Why other options are wrong:
- Option A: Incorrect. This wavelength does not correspond to the correct relationship between the orbit radius and the de Broglie wavelength for the 3rd orbit.
- Option C: Incorrect. This wavelength does not correspond to the correct relationship between the orbit radius and the de Broglie wavelength for the 3rd orbit.
- Option D: Incorrect. This wavelength does not correspond to the correct relationship between the orbit radius and the de Broglie wavelength for the 3rd orbit.

Key Concepts:
- Bohr's Theory
- de Broglie Wavelength
- Quantization of Angular Momentum

Memory Aids:
- Remember the relationship between the orbit radius and the de Broglie wavelength: $2\pi r = n\lambda$ .
- Remember that the radius of the nth Bohr orbit is proportional to $n^2$ .

Quick Tips:
- Always use the correct value of 'n' (orbit number) in the formulas.

128 Easy

128 of 176Question permalink

Understanding +4 | -1

The orbital cylindrically symmetrical about x-axis is:

View answer and explanation

Correct option: C

{p_x}

Option C (

{p_x}

) is correct because the

{p_x}

orbital is oriented along the x-axis and is cylindrically symmetrical about that axis.

Solution approach

Correct Answer + Core Reason: ${p_x}$ . The orbital cylindrically symmetrical about x-axis is ${p_x}$ .

Step-by-Step Solution:

Step 1: Define cylindrical symmetry: Cylindrical symmetry about the x-axis means that the orbital's shape is the same when rotated around the x-axis.

Step 2: Consider the p-orbitals: ${p_x}$ is oriented along the x-axis, ${p_y}$ is oriented along the y-axis, and ${p_z}$ is oriented along the z-axis.

Step 3: Determine the orbital with cylindrical symmetry about the x-axis: Only ${p_x}$ has cylindrical symmetry about the x-axis.

Step 4: Final Answer: The orbital cylindrically symmetrical about the x-axis is ${p_x}$ .

Why other options are wrong:
- Option A: ${p_z}$ is oriented along the z-axis and is not cylindrically symmetrical about the x-axis.
- Option B: ${p_y}$ is oriented along the y-axis and is not cylindrically symmetrical about the x-axis.
- Option D: ${d_{xz}}$ is not cylindrically symmetrical about the x-axis. It has two lobes in the xz-plane.

Key Concepts:
- Atomic orbitals
- p-orbitals ( ${p_x}$ , ${p_y}$ , ${p_z}$ )
- Cylindrical symmetry

Memory Aids:
- p-orbitals are dumbbell-shaped.

Quick Tips:
- Visualize the orientation of the orbitals in 3D space to determine their symmetry.

129 Easy

129 of 176Question permalink

Analyzing +4 | -1

Which statement relating to the spectrum of H atom is false?

View answer and explanation

Correct option: C

The spectral lines are closer together at longer wavelengths

Option C (The spectral lines are closer together at longer wavelengths) is false because spectral lines in the hydrogen atom spectrum are closer together at *shorter* wavelengths, converging towards the series limit.

Solution approach

Correct Answer + Core Reason: The spectral lines are closer together at longer wavelengths. This statement is false because the spectral lines converge at shorter wavelengths.

Step-by-Step Solution:

Step 1: Recall the Balmer series formula: $\frac{1}{\lambda} = R(\frac{1}{2^2} - \frac{1}{n^2})$ , where $R$ is the Rydberg constant and $n > 2$ .

Step 2: As $n$ increases, the difference between energy levels decreases, leading to smaller differences in wavelengths.

Step 3: This means the spectral lines get closer together as $n$ approaches infinity, which corresponds to shorter wavelengths, not longer wavelengths.

Step 4: Therefore, the statement that spectral lines are closer together at longer wavelengths is false.

Why other options are wrong:
- Option A: The lines can be defined by quantum number: This is true; each spectral line corresponds to a transition between specific energy levels defined by quantum numbers.
- Option B: The lines of longest wavelength in the Balmer series corresponds to the transition between n=3 and n=2 levels: This is true; the smallest energy difference (n=3 to n=2) corresponds to the longest wavelength.
- Option D: A continuum occurs at $n = \infty$ : This is true; at $n = \infty$ , the energy levels become continuous, leading to a continuum.

Key Concepts:
- Hydrogen atom spectrum
- Balmer series
- Rydberg constant
- Energy levels and transitions

Memory Aids:
- Remember that spectral lines converge at shorter wavelengths.

Quick Tips:
- Understand the relationship between energy levels, transitions, and wavelengths in the hydrogen atom spectrum.

130 Easy

130 of 176Question permalink

Analyzing +4 | -1

Which of the following pair is not isoelectronic species?

(At. no. Sm, 62; Er, 68; Yb, 70; Lu, 71; Eu, 63; Tb, 65 ; \mathrm{Tm}, 69)

View answer and explanation

Correct option: A

\mathrm{Sm}^{2+}

and

\mathrm{Er}^{3+}

Option A (

\mathrm{Sm}^{2+}

and

\mathrm{Er}^{3+}

) is correct because they do not have the same number of electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{Sm}^{2+}$ and $\mathrm{Er}^{3+}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each ion.

Step 2: $\mathrm{Sm}^{2+}$ : Samarium (Sm) has an atomic number of 62. $\mathrm{Sm}^{2+}$ loses 2 electrons, so it has 62 - 2 = 60 electrons.

Step 3: $\mathrm{Er}^{3+}$ : Erbium (Er) has an atomic number of 68. $\mathrm{Er}^{3+}$ loses 3 electrons, so it has 68 - 3 = 65 electrons.

Step 4: $\mathrm{Yb}^{2+}$ : Ytterbium (Yb) has an atomic number of 70. $\mathrm{Yb}^{2+}$ loses 2 electrons, so it has 70 - 2 = 68 electrons.

Step 5: $\mathrm{Lu}^{3+}$ : Lutetium (Lu) has an atomic number of 71. $\mathrm{Lu}^{3+}$ loses 3 electrons, so it has 71 - 3 = 68 electrons.

Step 6: $\mathrm{Eu}^{2+}$ : Europium (Eu) has an atomic number of 63. $\mathrm{Eu}^{2+}$ loses 2 electrons, so it has 63 - 2 = 61 electrons.

Step 7: $\mathrm{Tb}^{4+}$ : Terbium (Tb) has an atomic number of 65. $\mathrm{Tb}^{4+}$ loses 4 electrons, so it has 65 - 4 = 61 electrons.

Step 8: $\mathrm{Tm}^{4+}$ : Thulium (Tm) has an atomic number of 69. $\mathrm{Tm}^{4+}$ loses 4 electrons, so it has 69 - 4 = 65 electrons.

Step 9: Comparing the number of electrons, $\mathrm{Sm}^{2+}$ (60 electrons) and $\mathrm{Er}^{3+}$ (65 electrons) are not isoelectronic.

Step 10: Final Answer: $\mathrm{Sm}^{2+}$ and $\mathrm{Er}^{3+}$

Why other options are wrong:
- **Option B ( $\mathrm{Yb}^{2+}$ and $\mathrm{Lu}^{3+}$ ): Both have 68 electrons, so they are isoelectronic.
- Option C ( $\mathrm{Eu}^{2+}$ and $\mathrm{Tb}^{4+}$ ): Both have 61 electrons, so they are isoelectronic.
- Option D ( $\mathrm{Tb}^{4+}$ and $\mathrm{Tm}^{4+}$ ): Both have 65 electrons, so they are isoelectronic.

Key Concepts:
- Isoelectronic Species
- Atomic Number
- Ion Formation

Memory Aids:
- Iso- means same, so isoelectronic means same number of electrons.

Quick Tips:**
- Always calculate the number of electrons after the ion formation (loss or gain of electrons).

131 Easy

131 of 176Question permalink

Application +4 | -1

If there are five radial nodes, when what can be the correct representation of the orbital for

n = 8

View answer and explanation

Correct option: A

8d

The correct answer is 8d because the number of radial nodes is related to n and l by the formula n - l - 1, where l = 2 for d orbitals.

Solution approach

Correct Answer + Core Reason: 8d. The number of radial nodes is given by n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Step-by-Step Solution:

Step 1: Identify the given information: n = 8, number of radial nodes = 5.

Step 2: Use the formula for radial nodes: $n - l - 1 = \text{{number of radial nodes}}$

Step 3: Substitute the given values: $8 - l - 1 = 5$

Step 4: Solve for l: $7 - l = 5$ , so $l = 2$

Step 5: Determine the orbital corresponding to l = 2. l = 0 corresponds to s, l = 1 corresponds to p, l = 2 corresponds to d, l = 3 corresponds to f.

Step 6: Since l = 2, the orbital is d. Therefore, the correct representation is 8d.

Why other options are wrong:
- Option A (8d): This is the correct answer, as shown above.
- Option B (8s): For 8s, l = 0, so the number of radial nodes would be 8 - 0 - 1 = 7, not 5.
- Option C (8p): For 8p, l = 1, so the number of radial nodes would be 8 - 1 - 1 = 6, not 5.
- Option D (8f): For 8f, l = 3, so the number of radial nodes would be 8 - 3 - 1 = 4, not 5.

Key Concepts:
- Quantum numbers (n and l)
- Radial nodes
- Relationship between l and orbital type (s, p, d, f)

Memory Aids:
- Number of radial nodes = n - l - 1
- SPDF (acronym for the azimuthal quantum number l = 0, 1, 2, 3)

Quick Tips:
- Remember the relationship between l and the orbital type (s, p, d, f).
- Always use the correct formula for calculating radial nodes.

132 Easy

132 of 176Question permalink

Application +4 | -1

The ratio between the neutrons present in carbon and silicon with respect to atomic masses of 12 and 28 is:

View answer and explanation

Correct option: A

3:7

Option A is correct. Carbon-12 has 6 neutrons, and Silicon-28 has 14 neutrons, giving a ratio of 6:14, which simplifies to 3:7.

Solution approach

Correct Answer + Core Reason: $3:7$ . The number of neutrons is the difference between the mass number and the atomic number.

Step-by-Step Solution:

Step 1: Determine the number of neutrons in Carbon-12: Carbon has an atomic number of 6 and a mass number of 12. Number of neutrons = Mass number - Atomic number = $12 - 6 = 6$ .

Step 2: Determine the number of neutrons in Silicon-28: Silicon has an atomic number of 14 and a mass number of 28. Number of neutrons = Mass number - Atomic number = $28 - 14 = 14$ .

Step 3: Calculate the ratio of neutrons: $\frac{\text{Neutrons in Carbon}}{\text{Neutrons in Silicon}} = \frac{6}{14} = \frac{3}{7}$ .

Step 4: Express the ratio: The ratio is $3:7$ .

Why other options are wrong:
- Option B: $7:3$ : Incorrect; this is the inverse of the correct ratio.
- Option C: $3:4$ : Incorrect; this is not the correct simplified ratio of 6:14.
- Option D: $6:28$ : Incorrect; while these are the correct number of neutrons, the ratio is not simplified.

Key Concepts:
- Atomic number
- Mass number
- Neutrons

Memory Aids:
- Number of neutrons = Mass number - Atomic number.

Quick Tips:
- Always subtract the atomic number from the mass number to find the number of neutrons.

133 Easy

133 of 176Question permalink

Analyzing +4 | -1

Which of the following transitions of electrons in the H-atom will emit maximum energy

View answer and explanation

Correct option: C

{n_3} \to {n_2}

Option C (

{n_3} \to {n_2}

) is correct because the energy difference between n=3 and n=2 is the largest among the given transitions, resulting in the emission of maximum energy.

Solution approach

Correct Answer + Core Reason: ${n_3} \to {n_2}$ . This transition emits the maximum energy because the energy difference is the largest.

Step-by-Step Solution:

Step 1: Recall the energy level formula for the hydrogen atom: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Step 2: Calculate the energy difference for each transition:
- A: ${n_5} \to {n_4}$ : $\Delta E = -13.6(\frac{1}{5^2} - \frac{1}{4^2}) = -13.6(\frac{1}{25} - \frac{1}{16}) = 0.166 \text{ eV}$
- B: ${n_4} \to {n_3}$ : $\Delta E = -13.6(\frac{1}{4^2} - \frac{1}{3^2}) = -13.6(\frac{1}{16} - \frac{1}{9}) = 0.661 \text{ eV}$
- C: ${n_3} \to {n_2}$ : $\Delta E = -13.6(\frac{1}{3^2} - \frac{1}{2^2}) = -13.6(\frac{1}{9} - \frac{1}{4}) = 1.889 \text{ eV}$

Step 3: Compare the energy differences: The transition ${n_3} \to {n_2}$ has the largest energy difference.

Step 4: Final Answer: ${n_3} \to {n_2}$ emits the maximum energy.

Why other options are wrong:
- Option A: This transition has a smaller energy difference than ${n_3} \to {n_2}$ .
- Option B: This transition has a smaller energy difference than ${n_3} \to {n_2}$ .
- Option D: All will emit same energy: This is incorrect as the energy differences are different for each transition.

Key Concepts:
- Hydrogen atom energy levels
- Electronic transitions
- Energy difference and emitted energy

Memory Aids:
- Remember that larger energy level differences result in higher energy emissions.

Quick Tips:
- Calculate the energy difference for each transition to compare them accurately.

134 Easy

134 of 176Question permalink

Analyzing +4 | -1

Which one is wrong statement?

View answer and explanation

Correct option: C

The magnetic quantum number of an electron present in 2nd orbit can be ‘ –2’

Option C is the wrong statement because the magnetic quantum number (m) for an electron in the 2nd orbit (n=2) can only range from -1 to +1, including 0, and cannot be -2.

Solution approach

Correct Answer + Core Reason: The magnetic quantum number of an electron present in 2nd orbit can be ‘ –2’. This statement is wrong because for n=2, the possible values of l are 0 and 1, and the possible values of m are -l to +l, so m cannot be -2.

Step-by-Step Solution:

Step 1: Recall the rules for quantum numbers:
- Principal quantum number (n): n = 1, 2, 3, ...
- Azimuthal quantum number (l): l = 0, 1, 2, ..., (n-1)
- Magnetic quantum number (m): m = -l, -l+1, ..., 0, ..., l-1, l
- Spin quantum number (s): s = +1/2 or -1/2

Step 2: Analyze option A: Schrodinger wave equation cannot give information about spin quantum number. This is a correct statement because the Schrodinger equation does not explicitly include spin.

Step 3: Analyze option B: The orbital angular momentum of a s – electron in 3rd orbit zero. This is a correct statement because for an s-electron, l=0, and the orbital angular momentum is given by $\sqrt{{l(l+1)}} \frac{{h}}{{2\pi}} = 0$ .

Step 4: Analyze option C: The magnetic quantum number of an electron present in 2nd orbit can be ‘ –2’. For n=2, l can be 0 or 1. If l=0, m=0. If l=1, m can be -1, 0, or 1. Therefore, m cannot be -2. This statement is wrong.

Step 5: Analyze option D: In H – atom, the order of energies of orbitals is: 3s = 3p = 3d. This is a correct statement because in a hydrogen atom, the energy depends only on the principal quantum number (n).

Why other options are wrong:
- Option A: The Schrodinger equation does not directly account for electron spin, making this a correct statement.
- Option B: For an s-electron (l=0), the orbital angular momentum is zero, making this a correct statement.
- Option D: In a hydrogen atom, the energy levels depend only on the principal quantum number, so orbitals with the same n have the same energy, making this a correct statement.

Key Concepts:
- Quantum numbers (n, l, m, s)
- Schrodinger wave equation
- Orbital angular momentum
- Energy levels in hydrogen atom

Memory Aids:
- Remember the ranges for each quantum number: n > l, -l <= m <= l, s = +/- 1/2

Quick Tips:
- Always check the allowed values for each quantum number based on the other quantum numbers.
- Remember that in a hydrogen atom, energy levels depend only on n.

135 Easy

135 of 176Question permalink

Remembering +4 | -1

According to Bohr's postulates which quantity can take up only discrete values:

View answer and explanation

Correct option: B

Angular momentum

Option B (Angular momentum) is correct because, according to Bohr's postulates, the angular momentum of an electron is quantized and can only take discrete values.

Solution approach

Correct Answer + Core Reason: Angular momentum. According to Bohr's postulates, angular momentum is quantized.

Step-by-Step Solution:

Step 1: Recall Bohr's postulates.

Step 2: Bohr's postulates state that:
- Electrons revolve in specific orbits without emitting radiation.
- Electrons can only occupy orbits where their angular momentum is an integer multiple of $\frac{h}{2\pi}$ . That is, $L = n\frac{h}{2\pi}$ , where n is an integer.
- Electrons can jump from one orbit to another by absorbing or emitting energy equal to the energy difference between the orbits.

Step 3: Identify the quantity that can take only discrete values. According to Bohr's postulates, angular momentum can only take discrete values.

Why other options are wrong:
- Option A: Kinetic energy can take continuous values, although the energy levels are quantized.
- Option C: Momentum can take continuous values.
- Option D: Potential energy can take continuous values, although the energy levels are quantized.

Key Concepts:
- Bohr's postulates
- Quantization of angular momentum

Memory Aids:
- Remember that angular momentum is quantized in Bohr's model: $L = n\frac{h}{2\pi}$

Quick Tips:
- Recall Bohr's postulates to determine which quantities are quantized.

136 Easy

136 of 176Question permalink

Application +4 | -1

Which of the following pairs of nucleides are isodiaphers ?

View answer and explanation

Correct option: D

{ }_{25}^{55} \mathrm{Mn}

and

{ }_{30}^{65} \mathrm{Zn}

Option D (

{}_{25}^{55} \mathrm{Mn}

and

{}_{30}^{65} \mathrm{Zn}

) is correct because isodiaphers have the same difference between the number of neutrons and protons (N-Z).

Solution approach

Correct Answer + Core Reason: ${}_{25}^{55} \mathrm{Mn}$ and ${}_{30}^{65} \mathrm{Zn}$ . Isodiaphers have the same difference between the number of neutrons and protons (N-Z).

Step-by-Step Solution:

Step 1: Isodiaphers are nuclides that have the same isotopic number (neutron excess), which is the difference between the number of neutrons (N) and the number of protons (Z), i.e., N - Z.

Step 2: Calculate N - Z for each pair:
- A) ${ }_6^{13} \mathrm{C}$ : N = 13 - 6 = 7, N - Z = 7 - 6 = 1. ${ }_8^{16} \mathrm{O}$ : N = 16 - 8 = 8, N - Z = 8 - 8 = 0. Not isodiaphers.
- B) ${ }_1^1 \mathrm{H}$ : N = 1 - 1 = 0, N - Z = 0 - 1 = -1. ${ }_1^2 \mathrm{H}$ : N = 2 - 1 = 1, N - Z = 1 - 1 = 0. Not isodiaphers.
- C) ${ }_1^3 \mathrm{H}$ : N = 3 - 1 = 2, N - Z = 2 - 1 = 1. ${ }_2^4 \mathrm{He}$ : N = 4 - 2 = 2, N - Z = 2 - 2 = 0. Not isodiaphers.
- D) ${ }_{25}^{55} \mathrm{Mn}$ : N = 55 - 25 = 30, N - Z = 30 - 25 = 5. ${ }_{30}^{65} \mathrm{Zn}$ : N = 65 - 30 = 35, N - Z = 35 - 30 = 5. Isodiaphers.

Why other options are wrong:
- Option A: The difference between neutrons and protons is not the same for both nuclides.
- Option B: The difference between neutrons and protons is not the same for both nuclides.
- Option C: The difference between neutrons and protons is not the same for both nuclides.

Key Concepts:
- Nuclides
- Isotopes
- Isotones
- Isobars
- Isodiaphers

Memory Aids:
- Isodiaphers: Same difference between neutrons and protons (N-Z).

Quick Tips:
- Calculate N-Z for each nuclide and compare.

137 Easy

137 of 176Question permalink

Application +4 | -1

In H atom, the electron is de-excited from 5th shell to 1st shell. How many different lines may appear in line spectrum?

View answer and explanation

Correct option: C

10

Option C (10) is correct because the number of spectral lines when an electron de-excites from nth shell to the ground state is given by n(n-1)/2. For n=5, it is 5(5-1)/2 = 10.

Solution approach

Correct Answer + Core Reason: 10. The number of spectral lines is calculated using the formula n(n-1)/2.

Step-by-Step Solution:

Step 1: Identify the initial and final energy levels: initial level (n) = 5, final level = 1.

Step 2: Use the formula for the number of spectral lines: $N = \frac{{n(n-1)}}{2}$

Step 3: Substitute n = 5 into the formula: $N = \frac{{5(5-1)}}{2} = \frac{{5(4)}}{2} = \frac{{20}}{2} = 10$

Step 4: Final Answer: 10

Why other options are wrong:
- Option A: 4: This is incorrect. It might be the number of levels between 5 and 1 (5-1=4), but it's not the number of spectral lines.
- Option B: 8: This is incorrect and doesn't follow the correct formula.
- Option D: 12: This is incorrect and doesn't follow the correct formula.

Key Concepts:
- Hydrogen atom spectrum
- Number of spectral lines

Memory Aids:
- Remember the formula: Number of spectral lines = n(n-1)/2

Quick Tips:
- Always use the correct formula to calculate the number of spectral lines.

138 Easy

138 of 176Question permalink

Remembering +4 | -1

Orbitals processing the same energy are called:

View answer and explanation

Correct option: D

Degenerate orbitals

Option D is correct because orbitals possessing the same energy are called degenerate orbitals.

Solution approach

Correct Answer + Core Reason: Degenerate orbitals. Orbitals with the same energy are called degenerate orbitals.

Step-by-Step Solution:

Step 1: Define degenerate orbitals.
- Degenerate orbitals are atomic orbitals that have the same energy level. This typically occurs in hydrogen-like atoms (atoms with only one electron) or in the absence of external fields.

Step 2: Consider the other options.
- Hybrid orbitals are formed by mixing atomic orbitals.
- Valency orbitals are the orbitals involved in chemical bonding.
- d-orbitals are a specific type of atomic orbital with $l = 2$ .

Why other options are wrong:
- Option A: Hybrid orbitals are formed by mixing atomic orbitals, not orbitals with the same energy.
- Option B: Valency orbitals are involved in bonding, not necessarily having the same energy.
- Option C: d-orbitals are a specific type of orbital, not a general term for orbitals with the same energy.

Key Concepts:
- Degenerate orbitals
- Atomic orbitals

Memory Aids:
- Think of "degenerate" as meaning "equal in energy".

Quick Tips:
- Remember that degeneracy is lifted in multi-electron atoms due to electron-electron interactions.

139 Easy

139 of 176Question permalink

Application +4 | -1

An ion which has 18 electrons in the outermost shell is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{u}^ + }

Option A (

\mathrm{C}{\mathrm{u}^ + }

) is correct because Cu⁺ has the electronic configuration [Ar]3d¹⁰, which has 18 electrons in the outermost shell (n=3: 3s² 3p⁶ 3d¹⁰ = 2+6+10 = 18).

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{u}^ + }$ . Cu⁺ has 18 electrons in its outermost shell.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each ion.

Step 2: Identify the outermost shell for each ion.

Step 3: Count the number of electrons in the outermost shell.

Step 4: Cu⁺: [Ar] 3d¹⁰ (outermost shell: n=3, 3s² 3p⁶ 3d¹⁰ = 2+6+10 = 18 electrons)\nTh⁴⁺: [Rn] 5f⁰ 6d⁰ (outermost shell: n=6, 6s⁰ 6p⁰ = 0 electrons)\nCs⁺: [Xe] (outermost shell: n=5, 5s² 5p⁶ = 8 electrons)\nK⁺: [Ar] (outermost shell: n=3, 3s² 3p⁶ = 8 electrons)

Step 5: Identify the ion with 18 electrons in the outermost shell.

Why other options are wrong:
- Option B: $\mathrm{T}{\mathrm{h}^{4 + }}$ is incorrect because it has 0 electrons in its outermost shell.
- Option C: $\mathrm{C}{\mathrm{s}^ + }$ is incorrect because it has 8 electrons in its outermost shell.
- Option D: ${\mathrm{K}^ + }$ is incorrect because it has 8 electrons in its outermost shell.

Key Concepts:
- Electronic configuration
- Outermost shell

Memory Aids:
- Remember to remove electrons from the outermost shell first when forming cations.

Quick Tips:
- Always write the complete electronic configuration to accurately count the number of electrons in the outermost shell.

140 Easy

140 of 176Question permalink

Analyzing +4 | -1

Statement (S): Specific charge of

\alpha

-particles is twice to that of proton.

Explanation (E): Specific charge is given by

e / m.

View answer and explanation

Correct option: B

S is wrong but E is correct

Option B (S is wrong but E is correct) is the correct answer because the specific charge of an alpha particle is half that of a proton, and the explanation of specific charge being e/m is correct.

Solution approach

Correct Answer + Core Reason: S is wrong but E is correct. The specific charge of an alpha particle is half that of a proton, while the definition of specific charge is correct.

Step-by-Step Solution:

Step 1: Define specific charge. Specific charge is the ratio of charge to mass, given by $e/m$ .

Step 2: Consider the alpha particle. An alpha particle consists of 2 protons and 2 neutrons. Therefore, its charge is +2e and its mass is approximately 4 times the mass of a proton (4mp).

Step 3: Calculate the specific charge of an alpha particle. Specific charge of alpha particle = $\frac{{2e}}{{4m_p}} = \frac{{e}}{{2m_p}}$

Step 4: Calculate the specific charge of a proton. Specific charge of proton = $\frac{{e}}{{m_p}}$

Step 5: Compare the specific charges. The specific charge of the alpha particle is half that of the proton.

Step 6: Evaluate the statement (S). The statement (S) claims the specific charge of the alpha particle is twice that of the proton, which is incorrect.

Step 7: Evaluate the explanation (E). The explanation (E) correctly defines specific charge as $e/m$ .

Step 8: Conclusion. Statement (S) is wrong, and explanation (E) is correct.

Why other options are wrong:
- Option A: S is correct but E is wrong: This is incorrect because S is wrong and E is correct.
- Option C: Both S and E are correct and E is correct explanation of S: This is incorrect because S is wrong.
- Option D: Both S and E are correct but E is not correct explanation of S: This is incorrect because S is wrong.

Key Concepts:
- Specific charge
- Composition of alpha particles
- Charge and mass of proton

Memory Aids:
- Alpha particle: 2 protons + 2 neutrons (He nucleus)
- Specific charge = charge/mass

Quick Tips:
- Remember the composition of alpha particles to determine their charge and mass.
- Always define the terms before comparing them.

141 Easy

141 of 176Question permalink

Understanding +4 | -1

Dimensions of Planck’s constant are:

View answer and explanation

Correct option: D

energy × time

Option D (energy × time) is correct because Planck's constant (h) relates energy and frequency (E = hf), so h = E/f, and since frequency is inverse of time, h has dimensions of energy × time.

Solution approach

Correct Answer + Core Reason: energy × time. Planck's constant relates energy and frequency (E = hf).

Step-by-Step Solution:

Step 1: Recall the equation relating energy and frequency: $E = hf$ , where E is energy, h is Planck's constant, and f is frequency.

Step 2: Rearrange the equation to solve for Planck's constant: $h = E/f$ .

Step 3: Frequency is the inverse of time: $f = 1/T$ , where T is time.

Step 4: Substitute this into the equation for Planck's constant: $h = E \times T$ .

Step 5: Therefore, the dimensions of Planck's constant are energy × time.

Why other options are wrong:
- Option A (force × time): This represents impulse, not Planck's constant.
- Option B (energy × distance): This does not represent Planck's constant.
- Option C (energy/time): This represents power, not Planck's constant.

Key Concepts:
- Planck's constant
- Energy of a photon: $E = hf$
- Frequency and time period

Memory Aids:
- Remember the formula $E = hf$ .

Quick Tips:
- Know the relationship between energy, frequency, and Planck's constant.

142 Easy

142 of 176Question permalink

Understanding +4 | -1

When beryllium is bambarded with

\alpha

- particles extremely penetrating 396s which cannot be deflected by electrical or magnetic field are given out. These are

View answer and explanation

Correct option: C

A bean of neutrons

Option C (A bean of neutrons) is correct because when beryllium is bombarded with alpha particles, neutrons are emitted, which are neutral and highly penetrating.

Solution approach

Correct Answer + Core Reason: A bean of neutrons. Alpha particle bombardment of Beryllium produces neutrons.

Step-by-Step Solution:

Step 1: Recall the nuclear reaction when beryllium is bombarded with alpha particles.

Step 2: Write the nuclear reaction: $^9_4Be + ^4_2He \rightarrow ^{12}_6C + ^1_0n$

Step 3: Identify the emitted particle. The emitted particle is a neutron ( $^1_0n$ ).

Step 4: Recognize that neutrons are neutral and therefore not deflected by electric or magnetic fields.

Why other options are wrong:
- Option A: A beam of protons is deflected by electric and magnetic fields because protons are positively charged.
- Option B: $\alpha$ - rays are deflected by electric and magnetic fields because alpha particles are positively charged.
- Option D: $\beta$ rays are deflected by electric and magnetic fields because beta particles are negatively charged (electrons).

Key Concepts:
- Nuclear reactions
- Alpha particle bombardment
- Neutron emission

Memory Aids:
- Remember that alpha bombardment of beryllium is a common method for producing neutrons.

Quick Tips:
- Know the common nuclear reactions and the particles involved.

143 Easy

143 of 176Question permalink

Understanding +4 | -1

The electric configuration of element with atomic number 24 is

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because Chromium (Cr) with atomic number 24 has an exception to Hund's rule and Aufbau principle, where it achieves a more stable half-filled d-orbital configuration.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}$ . Chromium has an electronic configuration that deviates from the Aufbau principle due to the stability of half-filled d orbitals.

Step-by-Step Solution:

Step 1: Write the expected electronic configuration based on the Aufbau principle.

Step 2: The expected configuration would be $1s^2 2s^2 2p^6 3s^2 3p^6 3d^4 4s^2$ .

Step 3: Recognize that Chromium is an exception.

Step 4: Due to the stability of half-filled and fully filled d orbitals, one electron from the 4s orbital moves to the 3d orbital to achieve a half-filled configuration.

Step 5: The actual electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ .

Why other options are wrong:
- Option A: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^4},4{s^2}$ . This is the expected configuration based on the Aufbau principle, but it's incorrect for Chromium.
- Option B: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$ . This configuration is incorrect for Chromium.
- Option C: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6}$ . This configuration is incorrect for Chromium.

Key Concepts:
- Electronic Configuration
- Aufbau Principle
- Hund's Rule
- Stability of Half-Filled and Fully Filled Orbitals

Memory Aids:
- Chromium and Copper are exceptions to the Aufbau principle.

Quick Tips:
- Remember the exceptions to the Aufbau principle for elements like Chromium and Copper.

144 Easy

144 of 176Question permalink

Remembering +4 | -1

In the atomic spectrum of hydrogen the series of lines observed in the visible region is:

View answer and explanation

Correct option: A

Balmer series

Option A (Balmer series) is correct because the Balmer series corresponds to transitions to the

n=2

energy level, which results in emission lines in the visible region of the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: Balmer series. The Balmer series is in the visible region of the hydrogen spectrum.

Step-by-Step Solution:

Step 1: The question asks about the series of lines observed in the visible region of the hydrogen atomic spectrum.

Step 2: Recall the different series in the hydrogen spectrum:
- Lyman series: Ultraviolet region (transitions to $n=1$ )
- Balmer series: Visible region (transitions to $n=2$ )
- Paschen series: Infrared region (transitions to $n=3$ )
- Brackett series: Infrared region (transitions to $n=4$ )
- Pfund series: Infrared region (transitions to $n=5$ )

Step 3: Identify the series corresponding to the visible region.

Step 4: The Balmer series is the only one in the visible region.

Why other options are wrong:
- Option B (Paschen series): This is incorrect. The Paschen series is in the infrared region.
- Option C (Bracket series): This is incorrect. The Brackett series is in the infrared region.
- Option D (Lyman series): This is incorrect. The Lyman series is in the ultraviolet region.

Key Concepts:
- Hydrogen atomic spectrum
- Lyman series
- Balmer series
- Paschen series
- Brackett series

Memory Aids:
- Remember the order of the series and their corresponding regions of the electromagnetic spectrum: Lyman (UV), Balmer (Visible), Paschen (IR), Brackett (IR), Pfund (IR).

Quick Tips:
- Knowing the regions of the electromagnetic spectrum associated with each series is crucial for answering this type of question.

145 Easy

145 of 176Question permalink

Application +4 | -1

The volume of a proton is approximately;

View answer and explanation

Correct option: B

1.5 \times {10^{ - 38}}\mathrm{c}{\mathrm{m}^3}

Option B (

1.5 \times {10^{ - 38}}\mathrm{c}{\mathrm{m}^3}

) is correct because it is the approximate volume of a proton, calculated using its radius.

Solution approach

Correct Answer + Core Reason: $1.5 \times {10^{ - 38}}\mathrm{c}{\mathrm{m}^3}$ . This is the approximate volume of a proton.

Step-by-Step Solution:

Step 1: The radius of a proton is approximately $1 \times 10^{-13}$ cm.

Step 2: Assuming the proton is spherical, its volume can be calculated using the formula for the volume of a sphere:
$V = \frac{4}{3} \pi r^3$

Step 3: Substitute the radius into the formula:\nV = $\frac{4}{3}$ π (1 × 10^-13 cm)³ $\nV = \frac{4}{3} \pi (1 \times 10^{-39} \text{ cm}^3)$ \nV ≈ 4.19 × 10^-39 cm³ $\n\n**Step 4:** The closest option to this calculated value is$ 1.5 × 10^-38 cm³ $.\n\n**Why other options are wrong:**\n- **Option A:** Too large.\n- **Option C:** Too large.\n- **Option D:** The correct answer is one of the provided options.\n\n**Key Concepts:**\n- Proton radius\n- Volume of a sphere\n\n**Memory Aids:**\n- Remember the approximate radius of a proton:$ 10^-13 $cm.\n\n**Quick Tips:**\n- Know the approximate sizes of subatomic particles.$

146 Easy

146 of 176Question permalink

Understanding +4 | -1

The ratio of e/m,i.e., specific charge for a cathode ray:

View answer and explanation

Correct option: B

Is constant

Option B (Is constant) is correct because the specific charge (e/m) of a cathode ray (electron) is a fundamental constant and does not depend on the gas in the discharge tube or the cathode material.

Solution approach

Correct Answer + Core Reason: Is constant. The specific charge of a cathode ray is a fundamental constant.

Step-by-Step Solution:

Step 1: Define cathode rays.
- Cathode rays are streams of electrons.

Step 2: Define specific charge (e/m).
- Specific charge is the ratio of charge to mass.

Step 3: Explain why the specific charge of cathode rays is constant.
- Electrons are fundamental particles with a fixed charge and mass.
- Therefore, the e/m ratio is constant and does not depend on the gas or cathode material.

Why other options are wrong:
- Option A: Has the smallest value when the discharge tube is filled with ${H_2}$ is incorrect because the e/m ratio is constant.
- Option C: Varies with the atomic number of gas in the discharge tube is incorrect because the e/m ratio is constant.
- Option D: Varies with the atomic number of an element forming the cathode is incorrect because the e/m ratio is constant.

Key Concepts:
- Cathode rays (electrons)
- Specific charge (e/m)
- Fundamental constants

Memory Aids:
- Remember that electrons are fundamental particles with a fixed charge and mass.

Quick Tips:
- The specific charge is a characteristic property of a particle.

147 Easy

147 of 176Question permalink

Application +4 | -1

A certain radio station broadcasts on a frequency of 980 KHz, what is the wavelength of electromagnetic radiation broadcast by the radio station

View answer and explanation

Correct option: A

306 m

Option A (306 m) is correct because wavelength is calculated using the formula wavelength = speed of light / frequency.

Solution approach

Correct Answer + Core Reason: 306 m. Wavelength is calculated using the formula $\lambda = c/f$ .

Step-by-Step Solution:

Step 1: Identify the given values: Frequency (f) = 980 KHz = 980 × 10³ Hz, Speed of light (c) = 3 × 10⁸ m/s.

Step 2: Use the formula relating wavelength, frequency, and speed of light: $\lambda = \frac{c}{f}$

Step 3: Substitute the values: $\lambda = \frac{{3 \times 10^8 \text{{ m/s}}}}{{980 \times 10^3 \text{{ Hz}}}}$

Step 4: Calculate the wavelength: $\lambda = \frac{{3 \times 10^8}}{{980 \times 10^3}} \approx 306.12 \text{{ m}}$

Step 5: Round to the nearest whole number: $\lambda \approx 306 \text{{ m}}$

Why other options are wrong:
- Option B: This is too small. It might result from incorrectly using the frequency in MHz instead of Hz.
- Option C: This is also too small. It might result from a calculation error.
- Option D: This is too large. It might result from multiplying instead of dividing.

Key Concepts:
- Electromagnetic radiation
- Wavelength, frequency, and speed of light relationship: $c = \lambda f$

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

Quick Tips:
- Ensure the units are consistent before calculating. Convert KHz to Hz.

148 Easy

148 of 176Question permalink

Analyzing +4 | -1

Assertion : In case of isoelectronic ions the ionic size increases with the increase in atomic number.

Reason : The greater the attraction of nucleus, greater is the ionic radius

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D is correct because both the assertion and the reason are false. For isoelectronic ions, ionic size decreases with increasing atomic number, and greater nuclear attraction leads to a smaller ionic radius.

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. In isoelectronic species, ionic size decreases with increasing atomic number due to increased nuclear charge. Greater nuclear attraction leads to a smaller ionic radius.

Step-by-Step Solution:

Step 1: Analyze the assertion. For isoelectronic ions, the ionic size decreases as the atomic number increases. This is because the nuclear charge increases, pulling the electrons closer to the nucleus.

Step 2: Analyze the reason. Greater attraction of the nucleus leads to a smaller ionic radius, not a greater one.

Step 3: Conclude that both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both assertion and reason are false, so this option is incorrect.
- Option B: Both assertion and reason are false, so this option is incorrect.
- Option C: The assertion is false, so this option is incorrect.

Key Concepts:
- Isoelectronic species
- Ionic radius
- Nuclear charge

Memory Aids:
- More protons pull electrons closer, smaller size.

Quick Tips:
- Remember the trend of ionic size in isoelectronic species.

149 Easy

149 of 176Question permalink

Remembering +4 | -1

The atomic number of an element is derived from

View answer and explanation

Correct option: B

number of protons

Option B (number of protons) is correct because the atomic number uniquely identifies an element and is defined as the number of protons in the nucleus.

Solution approach

Correct Answer + Core Reason: number of protons. The atomic number is defined as the number of protons in the nucleus of an atom.

Step-by-Step Solution:

Step 1: Define Atomic Number: The atomic number (Z) of an element is the number of protons in the nucleus of an atom of that element.

Step 2: Relate Atomic Number to Element Identity: The atomic number uniquely identifies an element. For example, all atoms with an atomic number of 6 are carbon atoms.

Step 3: Consider the other options:
- Number of electrons: The number of electrons can vary in ions (charged atoms), so it's not a unique identifier.
- Number of neutrons: The number of neutrons can vary in isotopes of the same element.
- Number of isotopes: Isotopes are different forms of the same element with different numbers of neutrons.

Step 4: Conclusion: The atomic number is derived from the number of protons.

Why other options are wrong:
- Option A: The number of electrons can change when an atom becomes an ion, so it is not a defining characteristic of an element.
- Option C: The number of neutrons can vary in isotopes of the same element, so it is not a defining characteristic of an element.
- Option D: The number of isotopes is a property of an element, not the defining characteristic.

Key Concepts:
- Atomic Number
- Protons
- Neutrons
- Electrons
- Isotopes
- Ions

Memory Aids:
- Atomic Number = Number of Protons

Quick Tips:
- Remember that the atomic number defines the element.
- The number of protons is constant for a given element.

150 Easy

150 of 176Question permalink

Application +4 | -1

According to Bohr's theory, the angular momentum for an electron of 5th orbit is:

View answer and explanation

Correct option: A

\frac{{2.5h}}{\pi }

Option A (

\frac{{2.5h}}{\pi }

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

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

, where n is the orbit number.

Solution approach

Correct Answer + Core Reason: $\frac{{2.5h}}{\pi }$ . According to Bohr's theory, the angular momentum is quantized and given by $L = n\frac{h}{2\pi}$ .

Step-by-Step Solution:

Step 1: Recall Bohr's quantization condition for angular momentum:
$L = n \frac{h}{2\pi}$ , where:
$L$ is the angular momentum
$n$ is the principal quantum number (orbit number)
$h$ is Planck's constant

Step 2: Given that the electron is in the 5th orbit, $n = 5$ .

Step 3: Substitute $n = 5$ into the equation:
$L = 5 \frac{h}{2\pi} = \frac{5h}{2\pi} = \frac{2.5h}{\pi}$

Why other options are wrong:
- Option B: Incorrect. This would be the angular momentum for the 10th orbit, not the 5th.
- Option C: Incorrect. This value is not related to the Bohr's quantization condition for angular momentum.
- Option D: Incorrect. This would be the angular momentum for the 3rd orbit, not the 5th.

Key Concepts:
- Bohr's Theory
- Quantization of Angular Momentum

Memory Aids:
- Remember the formula $L = n \frac{h}{2\pi}$ for angular momentum in Bohr's theory.

Quick Tips:
- Always use the correct value of 'n' (orbit number) in the formula.

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