STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

Understanding +4 | -1

Which combinations of quantum numbers

n,l,m

and s for the electron in an atom does not provide a permissible solution of the wave equation?

View answer and explanation

Correct option: C

3,\;3,\;1,\; - \frac{1}{2}

Option C (

3,\;3,\;1,\; - \frac{1}{2}

) is correct because the azimuthal quantum number (l) cannot be equal to or greater than the principal quantum number (n). In this case, l=3 and n=3, which is not allowed.

Solution approach

Correct Answer + Core Reason: $3,\;3,\;1,\; - \frac{1}{2}$ . This combination of quantum numbers is not permissible because l cannot be equal to n.

Step-by-Step Solution:

Step 1: State the rules for quantum numbers:
- n (principal quantum number): n = 1, 2, 3, ...
- l (azimuthal quantum number): l = 0, 1, 2, ..., n-1
- ml (magnetic quantum number): ml = -l, -l+1, ..., 0, ..., l-1, l
- s (spin quantum number): s = +1/2 or -1/2

Step 2: Analyze each option:
- A: 3, 2, 1, 1/2 (n=3, l=2, ml=1, s=1/2) - Allowed (l < n, -l <= ml <= l)
- B: 3, 1, 1, -1/2 (n=3, l=1, ml=1, s=-1/2) - Allowed (l < n, -l <= ml <= l)
- C: 3, 3, 1, -1/2 (n=3, l=3, ml=1, s=-1/2) - Not allowed (l cannot be equal to n)
- D: 3, 2, -2, 1/2 (n=3, l=2, ml=-2, s=1/2) - Allowed (l < n, -l <= ml <= l)

Step 3: Final Answer: The combination 3, 3, 1, -1/2 is not a permissible solution because l cannot be equal to n.

Why other options are wrong:
- Option A: This combination is allowed because l < n and -l <= ml <= l.
- Option B: This combination is allowed because l < n and -l <= ml <= l.
- Option D: This combination is allowed because l < n and -l <= ml <= l.

Key Concepts:
- Quantum numbers (n, l, ml, s)
- Rules for quantum numbers

Memory Aids:
- Remember the allowed ranges for each quantum number.

Quick Tips:
- Always check if the quantum numbers satisfy the allowed ranges.

62 Medium

62 of 193Question permalink

Application +4 | -1

We can say that the energy of a photon of frequency v is given by

E = hv,

, where h is Planck’s constant. The momentum of a photon is

p = h/\lambda ,

, where λ is the wavelength of photon. Then we may conclude that velocity of light

I

equal to:

View answer and explanation

Correct option: B

E/p

Option B (

E/p

) is correct because the velocity of light can be expressed as the ratio of energy to momentum of a photon, derived from

E=hv

and

p=h/\lambda

.

Solution approach

Correct Answer + Core Reason: $E/p$ . The velocity of light is the ratio of energy to momentum.

Step-by-Step Solution:

Step 1: Given: $E = h\nu$ and $p = h/\lambda$

Step 2: We know that the velocity of light, $c = \nu \lambda$

Step 3: From $E = h\nu$ , we have $\nu = E/h$

Step 4: From $p = h/\lambda$ , we have $\lambda = h/p$

Step 5: Substitute $\nu$ and $\lambda$ into $c = \nu \lambda$ :
- $c = (E/h) \times (h/p) = E/p$

Step 6: Final Answer: $E/p$

Why other options are wrong:
- Option A: ${\left( {E/p} \right)^{1/2}}$ : Incorrect relationship between energy, momentum, and velocity.
- Option C: $Ep$ : Incorrect relationship between energy, momentum, and velocity.
- Option D: ${\left( {E/p} \right)^2}$ : Incorrect relationship between energy, momentum, and velocity.

Key Concepts:
- Energy of a photon: $E = h\nu$
- Momentum of a photon: $p = h/\lambda$
- Velocity of light: $c = \nu \lambda$

Memory Aids:
- Remember the relationships between energy, momentum, frequency, and wavelength.

Quick Tips:
- Use the fundamental relationships to derive the required expression.

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63 of 193Question permalink

Application +4 | -1

The momentum of radiation of wavelength 0.33 nm is …

\mathrm{kg\;m}{\sec ^{ - 1}}\;.

View answer and explanation

Correct option: A

2 \times {10^{ - 24}}

Option A (

2 \times {10^{ - 24}}

) is correct because momentum is calculated as h/λ.

Solution approach

Correct Answer + Core Reason: $2 \times {10^{ - 24}}$ . The momentum of radiation is given by $p = \frac{h}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength, $\lambda = 0.33 nm = 0.33 \times 10^{-9} m$ .\nWe need to find the momentum, $p$ .

Step 2: Use the formula $p = \frac{h}{\lambda}$ , where $h = 6.626 \times 10^{-34} Js$ .

Step 3: Substitute the values: $p = \frac{{6.626 \times 10^{-34} Js}}{{0.33 \times 10^{-9} m}}$

Step 4: Calculate the momentum: $p = 2.0078 \times 10^{-24} kg m/s \approx 2 \times 10^{-24} kg m/s$

Step 5: Final Answer: $p = 2 \times 10^{-24} kg m/s$

Why other options are wrong:
- Option B: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option C: This is incorrect. It might be obtained by using an incorrect formula or calculation.
- Option D: This is incorrect. It might be obtained by using an incorrect formula or calculation.

Key Concepts:
- Momentum of radiation
- Relationship between momentum and wavelength

Memory Aids:
- Remember the formula $p = \frac{h}{\lambda}$

Quick Tips:
- Convert all units to SI units before performing calculations.

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Analyzing +4 | -1

Which set has the same number of s-electrons?

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{u}^{2 + }},\mathrm{F}{\mathrm{e}^{2 + }},\mathrm{N}{\mathrm{i}^{2 + }}

Option B (Cu²⁺, Fe²⁺, Ni²⁺) is correct because each ion has 16 s-electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{u}^{2 + }},\mathrm{F}{\mathrm{e}^{2 + }},\mathrm{N}{\mathrm{i}^{2 + }}$ . These ions have the same number of s-electrons.

Step-by-Step Solution:

Step 1: Write the electronic configurations for each species.\nC: 1s² 2s² 2p² (4 s-electrons)\nCu²⁺: [Ar] 3d⁹ (1s² 2s² 2p⁶ 3s² 3p⁶) (16 s-electrons)\nZn: [Ar] 4s² (1s² 2s² 2p⁶ 3s² 3p⁶ 4s²) (18 s-electrons)\nFe²⁺: [Ar] 3d⁶ (1s² 2s² 2p⁶ 3s² 3p⁶) (16 s-electrons)\nNi²⁺: [Ar] 3d⁸ (1s² 2s² 2p⁶ 3s² 3p⁶) (16 s-electrons)\nS²⁻: [Ar] (1s² 2s² 2p⁶ 3s² 3p⁶) (18 s-electrons)

Step 2: Count the number of s-electrons in each species.

Step 3: Compare the number of s-electrons in each set.\nOnly Cu²⁺, Fe²⁺, and Ni²⁺ have the same number of s-electrons (16).

Why other options are wrong:
- Option A: C, Cu²⁺, and Zn have different numbers of s-electrons (4, 16, and 18 respectively).
- Option C: S²⁻, Ni²⁺, and Zn have different numbers of s-electrons (18, 16, and 18 respectively).
- Option D: The correct answer is B, so none of these is not the correct answer.

Key Concepts:
- Electronic configuration
- s-electrons

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

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

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Remembering +4 | -1

The metal mainly used in devising photoelectric cells is

View answer and explanation

Correct option: C

Cs

Option C (

Cs

) is correct because Cesium has a low ionization energy and is therefore commonly used in photoelectric cells.

Solution approach

Correct Answer + Core Reason: $Cs$ . Cesium is mainly used in photoelectric cells due to its low ionization energy.

Step-by-Step Solution:

Step 1: Recall the principle of photoelectric cells.

Step 2: Photoelectric cells require a material that easily emits electrons when exposed to light.

Step 3: Elements with low ionization energies are suitable for this purpose.

Step 4: Cesium (Cs) has a very low ionization energy compared to Na, Li, and Rb.

Why other options are wrong:
- Option A: $Na$ : Sodium has a higher ionization energy than Cesium.
- Option B: $Li$ : Lithium has a higher ionization energy than Cesium.
- Option D: $Rb$ : Rubidium has a lower ionization energy than Sodium and Lithium, but still higher than Cesium. Cesium is preferred.

Key Concepts:
- Photoelectric effect
- Ionization energy

Memory Aids:
- Cesium (Cs) = low ionization energy = photoelectric cells

Quick Tips:
- Remember that elements with low ionization energies are suitable for photoelectric cells.

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Understanding +4 | -1

Statement I: Photoelectric effect is most readily shown by cesium metal.
Statement II: Photoelectric effect is more pronounced in metals having high ionisation energy.

View answer and explanation

Correct option: D

Statement I is true but statement II is false

Option D is correct. Cesium readily shows the photoelectric effect due to its low ionization energy, while high ionization energy hinders the effect.

Solution approach

Correct Answer + Core Reason: Statement I is true but statement II is false. Cesium has low ionization energy, making it ideal for the photoelectric effect, while high ionization energy hinders it.

Step-by-Step Solution:

Step 1: Understand the photoelectric effect: It involves the emission of electrons when light shines on a material.

Step 2: Consider ionization energy: Low ionization energy means electrons are easily removed.

Step 3: Cesium's properties: Cesium has a low ionization energy, making it suitable for the photoelectric effect.

Step 4: Statement I analysis: Photoelectric effect is readily shown by cesium metal, which is true.

Step 5: Statement II analysis: Photoelectric effect is more pronounced in metals having high ionization energy, which is false.

Step 6: Conclusion: Statement I is true, and Statement II is false.

Why other options are wrong:
- Option A: Statement I is false but statement II is true: Incorrect because Statement I is true and Statement II is false.
- Option B: Both Statement I and statement II are true: Incorrect because Statement II is false.
- Option C: Both Statement I and statement II are false: Incorrect because Statement I is true.

Key Concepts:
- Photoelectric effect
- Ionization energy

Memory Aids:
- Low ionization energy facilitates electron emission.

Quick Tips:
- Remember the relationship between ionization energy and the photoelectric effect.

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Analyzing +4 | -1

In which one of the following pairs the two species are both isoelectronic and isotopic?
(Atomic numbers :

{Ca}=20, {Ar}=18, {~K}=19, {Mg}=12

,

{Fe}=26, {Na}=11

)

View answer and explanation

Correct option: B

{ }^{39} {~K}^{+}

and

{ }^{40} {~K}^{+}

Option B (

{ }^{39} {~K}^{+}

and

{ }^{40} {~K}^{+}

) is correct because both species have the same number of electrons (isoelectronic) and are isotopes of potassium (isotopic).

Solution approach

Correct Answer + Core Reason: ${ }^{39} {~K}^{+}$ and ${ }^{40} {~K}^{+}$ are both isoelectronic and isotopic.

Step-by-Step Solution:

Step 1: Define isoelectronic and isotopic.\nIsoelectronic: Having the same number of electrons.\nIsotopic: Having the same number of protons but different number of neutrons.

Step 2: Analyze Option A: ${ }^{40} {Ca}^{2+}$ and ${ }^{40} {Ar}$
${ }^{40} {Ca}^{2+}$ has 20 - 2 = 18 electrons.
${ }^{40} {Ar}$ has 18 electrons.\nThey are isoelectronic. However, they are different elements, so not isotopic.

Step 3: Analyze Option B: ${ }^{39} {~K}^{+}$ and ${ }^{40} {~K}^{+}$
${ }^{39} {~K}^{+}$ has 19 - 1 = 18 electrons.
${ }^{40} {~K}^{+}$ has 19 - 1 = 18 electrons.\nThey are isoelectronic. They are both isotopes of Potassium.

Step 4: Analyze Option C: ${ }^{24} {Mg}^{2+}$ and ${ }^{25} {Mg}$
${ }^{24} {Mg}^{2+}$ has 12 - 2 = 10 electrons.
${ }^{25} {Mg}$ has 12 electrons.\nThey are not isoelectronic.

Step 5: Analyze Option D: ${ }^{23} {Na}$ and ${ }^{24} {Na}^{+}$
${ }^{23} {Na}$ has 11 electrons.
${ }^{24} {Na}^{+}$ has 11 - 1 = 10 electrons.\nThey are not isoelectronic.

Step 6: Conclusion: Only Option B satisfies both conditions.

Why other options are wrong:
- Option A: ${ }^{40} {Ca}^{2+}$ and ${ }^{40} {Ar}$ are isoelectronic but not isotopes of the same element.
- Option C: ${ }^{24} {Mg}^{2+}$ and ${ }^{25} {Mg}$ are not isoelectronic.
- Option D: ${ }^{23} {Na}$ and ${ }^{24} {Na}^{+}$ are not isoelectronic.

Key Concepts:
- Isoelectronic species
- Isotopes

Memory Aids:
- Isoelectronic: \"iso-\" means same, \"electronic\" refers to electrons.
- Isotopes: Same number of protons, different number of neutrons.

Quick Tips:
- Calculate the number of electrons for each species carefully.
- Remember that ions gain or lose electrons.

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Remembering +4 | -1

The possibility of finding an electron in an orbital was conceived by:

View answer and explanation

Correct option: D

Schrödinger

Option D (Schrödinger) is correct because Erwin Schrödinger developed the wave equation that describes the probability of finding an electron in an orbital.

Solution approach

Correct Answer + Core Reason: Schrödinger. Schrödinger developed the wave equation that describes the probability of finding an electron.

Step-by-Step Solution:

Step 1: Recall the contributions of different scientists to the development of quantum mechanics.

Step 2: Rutherford proposed the nuclear model of the atom.

Step 3: Bohr developed a model of the hydrogen atom with quantized energy levels.

Step 4: Heisenberg formulated the uncertainty principle.

Step 5: Schrödinger developed the wave equation that describes the probability of finding an electron in an orbital.

Step 6: Therefore, the possibility of finding an electron in an orbital was conceived by Schrödinger.

Why other options are wrong:
- Option A: Rutherford: Rutherford's model focused on the nucleus, not the probability of finding electrons in orbitals.
- Option B: Bohr: Bohr's model quantized energy levels but didn't fully address the probability of electron location.
- Option C: Heisenberg: Heisenberg's uncertainty principle deals with the limits of knowing both position and momentum, but Schrödinger's equation directly addresses the probability of electron location.

Key Concepts:
- Quantum mechanics
- Schrödinger wave equation
- Atomic orbitals
- Probability density

Memory Aids:
- Remember that Schrödinger's equation is central to understanding the probability of finding electrons in atoms.

Quick Tips:
- Know the key contributions of different scientists to the development of atomic theory and quantum mechanics.

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Understanding +4 | -1

The number of electron ejected in the photoelectric experiment is proportional to the

View answer and explanation

Correct option: C

Both (1) and (2)

Option C is correct. Both intensity and brightness of light are directly proportional to the number of electrons ejected in the photoelectric effect.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). The number of electrons ejected is proportional to both the intensity and brightness of light.

Step-by-Step Solution:

Step 1: Define intensity: Intensity is the power per unit area of the light.

Step 2: Define brightness: Brightness is a subjective perception of intensity.

Step 3: Relate intensity and brightness to the number of electrons: Higher intensity/brightness means more photons, leading to more electrons being ejected.

Step 4: Conclusion: The number of electrons ejected is proportional to both intensity and brightness.

Why other options are wrong:
- Option A: Intensity of light: Incorrect because brightness is also proportional to the number of electrons ejected.
- Option B: Brightness of light: Incorrect because intensity is also proportional to the number of electrons ejected.
- Option D: None of the above: Incorrect because both intensity and brightness are proportional to the number of electrons ejected.

Key Concepts:
- Photoelectric effect
- Intensity of light
- Brightness of light

Memory Aids:
- More light (intensity/brightness) = more electrons.

Quick Tips:
- Remember that intensity and brightness are related and both affect the number of electrons ejected.

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Understanding +4 | -1

In photoelectric effect, the photo-current:

View answer and explanation

Correct option: C

Does not depend on the frequency of photon but depends only on the intensity of incident light

Option C (Does not depend on the frequency of photon but depends only on the intensity of incident light) is correct because the photocurrent is directly proportional to the number of photoelectrons emitted, which is determined by the intensity of the incident light.

Solution approach

Correct Answer + Core Reason: Does not depend on the frequency of photon but depends only on the intensity of incident light. Photocurrent depends on the number of photoelectrons emitted, which is determined by intensity.

Step-by-Step Solution:

Step 1: Define photocurrent.
- Photocurrent is the electric current produced by photoelectrons emitted in the photoelectric effect.

Step 2: Explain the relationship between photocurrent, intensity, and frequency.
- The intensity of incident light is proportional to the number of photons incident on the metal surface per unit time.
- The number of photoelectrons emitted is directly proportional to the number of incident photons (intensity).
- Therefore, the photocurrent is directly proportional to the intensity of the incident light.
- The frequency of the incident light affects the kinetic energy of the emitted electrons, but not the number of electrons (photocurrent), as long as the frequency is above the threshold frequency.

Why other options are wrong:
- Option A: Increases with increase of frequency of incident photon is incorrect because frequency affects the kinetic energy of the electrons, not the photocurrent.
- Option B: Decreases with increase of frequency of incident photon is incorrect because frequency does not directly affect the photocurrent.
- Option D: Depends both on intensity and frequency of the incident photon is incorrect because the primary dependence is on intensity.

Key Concepts:
- Photoelectric effect
- Photocurrent
- Intensity of light
- Frequency of light

Memory Aids:
- Remember that intensity affects the number of electrons, while frequency affects the energy of each electron.

Quick Tips:
- Photocurrent is directly proportional to intensity, assuming the frequency is above the threshold frequency.

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Analyzing +4 | -1

Assertion (A) : Element formed by the successive emission of

1 \alpha

-particle and

2 \beta

-particles is placed in the same group as the parent element.

Reason (R):

\alpha

-particle is

\left[{ }_2^4 \mathrm{He}\right]^{2+}

while

\beta

-particle is

{ }_{-1}^0 e

.

View answer and explanation

Correct option: B

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

.

Option B (Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

.) is correct because the position in the periodic table depends on the atomic number and the electronic configuration, not directly on the nature of alpha and beta particles.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . The group position depends on the atomic number after emission, and both statements are factually correct.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Emission of 1 alpha particle reduces the atomic number by 2, and emission of 2 beta particles increases the atomic number by 2. Therefore, the net change in atomic number is 0, and the element remains in the same group.

Step 2: Analyze Reason (R): $\alpha$ -particle is $\left[{ }_2^4 \mathrm{He}\right]^{2+}$ while $\beta$ -particle is ${ }_{-1}^0 e$ . This statement is correct as it defines the composition of alpha and beta particles.

Step 3: Determine if R explains A: While R is true, it doesn't directly explain why the element remains in the same group. The reason is the net change in atomic number being zero.

Why other options are wrong:
- Option A: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ .: This is incorrect because R doesn't directly explain A.
- Option C: $\mathbf{A}$ is true but $\mathbf{R}$ is false.: This is incorrect because both A and R are true.
- Option D: $\mathbf{A}$ is false but $\mathbf{R}$ is true.: This is incorrect because A is also true.

Key Concepts:
- Alpha decay
- Beta decay
- Atomic number
- Group in the periodic table

Memory Aids:
- Alpha decay reduces atomic number by 2, mass number by 4.
- Beta decay increases atomic number by 1, mass number remains the same.

Quick Tips:
- Track the changes in atomic number and mass number during radioactive decay to determine the new element's position in the periodic table.

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Remembering +4 | -1

Splitting of spherical lines when atoms are subjected to strong electric field is called:

View answer and explanation

Correct option: B

Stark effect

Option B (Stark effect) is correct because the Stark effect describes the splitting of spectral lines due to an external electric field.

Solution approach

Correct Answer + Core Reason: Stark effect. The Stark effect is the splitting of spectral lines due to an external electric field.

Step-by-Step Solution:

Step 1: The question asks about the splitting of spectral lines when atoms are subjected to a strong electric field.

Step 2: Recall the definitions of the Zeeman and Stark effects.

Step 3: The Zeeman effect is the splitting of spectral lines due to an external magnetic field. The Stark effect is the splitting of spectral lines due to an external electric field.

Step 4: Since the question refers to an electric field, the answer is the Stark effect.

Why other options are wrong:
- Option A: Zeeman effect is incorrect because it refers to the splitting of spectral lines due to a magnetic field, not an electric field.
- Option C: Decay is incorrect because it refers to radioactive decay, which is not related to the splitting of spectral lines due to an electric field.
- Option D: Disintegration is incorrect because it refers to the breaking apart of a nucleus, which is not related to the splitting of spectral lines due to an electric field.

Key Concepts:
- Stark effect
- Zeeman effect

Memory Aids:
- Stark = Electric, Zeeman = Magnetic

Quick Tips:
- Remember the definitions of the Stark and Zeeman effects.

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Analyzing +4 | -1

Which of the following is correctly matched?

View answer and explanation

Correct option: A

Momentum of H atom when electrons return from n=2 to n=1:

\frac{{3Rh}}{4}

Option A (Momentum of H atom when electrons return from n=2 to n=1:

\frac{{3Rh}}{{4}}

) is correct because the momentum of the emitted photon equals the momentum of the recoiling H atom.

Solution approach

Correct Answer + Core Reason: Momentum of H atom when electrons return from n=2 to n=1: $\frac{{3Rh}}{{4}}$ . The momentum of the emitted photon equals the momentum of the recoiling H atom.

Step-by-Step Solution:

Step 1: Calculate the wavelength of the photon emitted when the electron transitions from n=2 to n=1 in a hydrogen atom.

Step 2: Use the Rydberg formula: $\frac{1}{\lambda} = R(\frac{1}{n_1^2} - \frac{1}{n_2^2})$ , where R is the Rydberg constant.

Step 3: Substitute $n_1 = 1$ and $n_2 = 2$ : $\frac{1}{\lambda} = R(\frac{1}{1^2} - \frac{1}{2^2}) = R(1 - \frac{1}{4}) = \frac{3R}{4}$ .

Step 4: Therefore, $\lambda = \frac{4}{3R}$ .

Step 5: Calculate the momentum of the photon using the formula $p = \frac{h}{\lambda}$ , where h is Planck's constant.

Step 6: Substitute the value of $\lambda$ : $p = \frac{h}{\frac{4}{3R}} = \frac{3Rh}{4}$ .

Step 7: Since the momentum of the emitted photon is equal to the momentum of the recoiling hydrogen atom, the momentum of the H atom is also $\frac{3Rh}{4}$ .

Why other options are wrong:
- Option B: Momentum of photon is dependent on the wavelength of light, $p = h/\lambda$ .
- Option C: e/m ratio of anode rays depends on the gas in the discharge tube because anode rays are positive ions of the gas.
- Option D: Radius of nucleus is proportional to ${\left( {\mathrm{Mass\;no}.} \right)^{1/3}}$ , not ${\left( {\mathrm{Mass\;no}.} \right)^{1/2}}$ .

Key Concepts:
- Rydberg Formula
- Momentum of a Photon
- Conservation of Momentum

Memory Aids:
- Remember the Rydberg formula for calculating wavelengths of emitted photons.
- Momentum of a photon: $p = h/\lambda$ .

Quick Tips:
- The momentum of the emitted photon is equal in magnitude and opposite in direction to the momentum of the recoiling atom.

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Application +4 | -1

The mass of an electron is m, its charge is e and it is accelerated from rest through a potential difference, V. The velocity of electron will be calculated by formula

View answer and explanation

Correct option: C

\sqrt {\left( {\frac{{2eV}}{m}} \right)}

Option C (

\sqrt {\left( {\frac{{2eV}}{m}} \right)}

) is correct because it is derived from equating the kinetic energy gained by the electron to the work done by the electric field.

Solution approach

Correct Answer + Core Reason: $\sqrt {\left( {\frac{{2eV}}{m}} \right)}$ . The velocity is obtained by equating the kinetic energy gained to the work done by the potential difference.

Step-by-Step Solution:

Step 1: The electron is accelerated through a potential difference V.

Step 2: The work done on the electron is given by $W = eV$ , where e is the charge of the electron.

Step 3: The kinetic energy gained by the electron is given by $KE = \frac{1}{2}mv^2$ , where m is the mass of the electron and v is its velocity.

Step 4: Equate the work done to the kinetic energy gained: $eV = \frac{1}{2}mv^2$

Step 5: Solve for the velocity v: $v^2 = \frac{2eV}{m}$ , so $v = \sqrt{\frac{2eV}{m}}$

Why other options are wrong:
- **Option A ( $\sqrt {\frac{V}{m}}$ ): This is incorrect because it doesn't account for the charge of the electron.
- Option B ( $\sqrt {\frac{{eV}}{m}}$ ): This is incorrect because it's missing the factor of 2 in the numerator.
- Option D (None of these): This is incorrect because option C is the correct formula.

Key Concepts:
- Work-energy theorem
- Kinetic energy
- Electric potential energy

Memory Aids:
- Work done by electric field: W = qV
- Kinetic energy: KE = (1/2)mv²

Quick Tips:**
- Remember the relationship between work done by an electric field and the potential difference.
- Equate the work done to the kinetic energy gained to find the velocity.

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Application +4 | -1

What is the energy (in eV) require to excite the electron from n=1 to n=2 state in hydrogen atom?
(n=principle quantum number)

View answer and explanation

Correct option: D

10.2

Option D (10.2) is correct because it represents the energy difference between the n=1 and n=2 energy levels in a hydrogen atom.

Solution approach

Correct Answer + Core Reason: 10.2. The energy required to excite the electron from n=1 to n=2 is the difference in energy levels.

Step-by-Step Solution:

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

Step 2: Calculate the energy of the electron in the n=1 state: $E_1 = -\frac{13.6}{1^2} = -13.6$ eV.

Step 3: Calculate the energy of the electron in the n=2 state: $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4$ eV.

Step 4: The energy required to excite the electron from n=1 to n=2 is the difference in energy levels: $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = -3.4 + 13.6 = 10.2$ eV.

Why other options are wrong:
- Option A (13.6): This is the ionization energy of hydrogen, not the energy required for the n=1 to n=2 transition.
- Option B (3.4): This is the energy of the electron in the n=2 state (with a negative sign), not the energy difference.
- Option C (17): This is incorrect. It's likely a random value.

Key Concepts:
- Energy levels in hydrogen atom
- Bohr's model
- Energy difference between levels

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

Quick Tips:
- Remember the formula for the energy levels in a hydrogen atom.
- Calculate the energy difference between the initial and final states to find the energy required for the transition.

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76 of 193Question permalink

Remembering +4 | -1

The product of which of the following is equal to the velocity of light

View answer and explanation

Correct option: B

Wavelength and frequency

Option B (Wavelength and frequency) is correct because the product of wavelength and frequency equals the speed of light (

c = \lambda v

).

Solution approach

Correct Answer + Core Reason: Wavelength and frequency. The product of wavelength and frequency is equal to the velocity of light.

Step-by-Step Solution:

Step 1: Wave Equation: The relationship between the velocity of a wave (v), its wavelength ( $\lambda$ ), and its frequency (f or v) is given by the equation:\nv = λ v $\n\n**Step 2:** Velocity of Light: For electromagnetic waves, such as light, the velocity is the speed of light (c):\nc = \lambda v$ \nTherefore, the product of wavelength and frequency is equal to the velocity of light.

Why other options are wrong:
- Option A: Wavelength of wave number. Wave number is the reciprocal of wavelength, so their product is dimensionless and not equal to the velocity of light.
- Option C: Frequency and wave number. The product of frequency and wave number does not equal the velocity of light.
- Option D: Wavelength and amplitude. Amplitude is related to the intensity of the wave, not its velocity.

Key Concepts:
- Wave equation
- Wavelength
- Frequency
- Speed of light

Memory Aids:
- $c = \lambda v$ (Speed of light = Wavelength x Frequency).

Quick Tips:
- Remember the fundamental relationship between speed, wavelength, and frequency for waves.

77 Medium

77 of 193Question permalink

Application +4 | -1

Number of electrons in 1.8 mL of

{\mathrm{H}_2}\mathrm{O}

are:

View answer and explanation

Correct option: A

6.02 \times {10^{23}}

Option A (

6.02 \times {10^{23}}

) is correct because it represents the number of water molecules in 1.8 mL of water multiplied by the number of electrons in each water molecule.

Solution approach

Correct Answer + Core Reason: $6.02 \times {10^{23}}$ . This is the number of electrons in 1.8 mL of water.

Step-by-Step Solution:

Step 1: Determine the number of moles of water in 1.8 mL. Assuming the density of water is 1 g/mL, 1.8 mL of water has a mass of 1.8 g.

Step 2: Calculate the number of moles: $\text{{moles}} = \frac{{\text{{mass}}}}{{\text{{molar mass}}}} = \frac{{1.8 \text{{ g}}}}{{18 \text{{ g/mol}}}} = 0.1 \text{{ mol}}$

Step 3: Calculate the number of water molecules: $\text{{number of molecules}} = \text{{moles}} \times N_A = 0.1 \text{{ mol}} \times 6.022 \times 10^{23} \text{{ molecules/mol}} = 6.022 \times 10^{22} \text{{ molecules}}$

Step 4: Determine the number of electrons in one water molecule. Hydrogen has 1 electron, and oxygen has 8 electrons. So, ${\mathrm{H}_2}\mathrm{O}$ has $2(1) + 8 = 10$ electrons.

Step 5: Calculate the total number of electrons: $\text{{total electrons}} = \text{{number of molecules}} \times \text{{electrons per molecule}} = 6.022 \times 10^{22} \times 10 = 6.022 \times 10^{23} \text{{ electrons}}$

Why other options are wrong:
- **Option B ( $6.02 \times {10^{24}}$ ): This is incorrect because it's 10 times larger than the actual number of electrons. It might result from multiplying the number of moles by Avogadro's number and the number of electrons per molecule without correctly accounting for the initial volume.
- Option C ( $6.02 \times {10^{22}}$ ): This is the number of water molecules, not the number of electrons. It neglects the 10 electrons per water molecule.
- Option D ( $6.02 \times {10^{25}}$ ): This is incorrect and significantly larger than the actual number of electrons. It likely involves a miscalculation or misunderstanding of the powers of 10.

Key Concepts:
- Mole concept
- Avogadro's number
- Molar mass

Memory Aids:**
- $\text{{Moles}} = \frac{{\text{{Mass}}}}{{\text{{Molar Mass}}}}$
- $N_A = 6.022 \times 10^{23}$

Quick Tips:
- Always convert volume to mass using density if needed.
- Remember to multiply the number of molecules by the number of atoms/electrons per molecule.

78 Medium

78 of 193Question permalink

Understanding +4 | -1

The ratio of the radius of the orbit for the electron orbiting the hydrogen nucleus to that of an electron orbiting a deuterium nucleus is:

View answer and explanation

Correct option: A

1∶1

Option A (1∶1) is correct because the radius of the orbit depends on the principal quantum number and atomic number, not the mass of the nucleus.

Solution approach

Correct Answer + Core Reason: 1∶1. The radius of the orbit depends on the principal quantum number and atomic number, not the mass of the nucleus.

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth orbit in a hydrogen-like atom: $r_n = \frac{{n^2 a_0}}{Z}$ , where $n$ is the principal quantum number, $a_0$ is the Bohr radius, and $Z$ is the atomic number.

Step 2: Analyze the given information: We are comparing the radius of the orbit for an electron orbiting a hydrogen nucleus (H) to that of an electron orbiting a deuterium nucleus (D). Both are assumed to be in the same orbit (same $n$ ).

Step 3: Determine the atomic numbers: For hydrogen (H), $Z = 1$ . For deuterium (D), $Z = 1$ (Deuterium is an isotope of hydrogen, so it has the same atomic number).

Step 4: Calculate the ratio of the radii: $\frac{{r_H}}{{r_D}} = \frac{{\frac{{n^2 a_0}}{{Z_H}}}}{{\frac{{n^2 a_0}}{{Z_D}}}} = \frac{{Z_D}}{{Z_H}} = \frac{1}{1} = 1$

Step 5: Final Answer: 1∶1

Why other options are wrong:
- Option B: 1∶2: Incorrect, the mass of the nucleus does not affect the radius.
- Option C: 2∶1: Incorrect, the mass of the nucleus does not affect the radius.
- Option D: 1∶3: Incorrect, the mass of the nucleus does not affect the radius.

Key Concepts:
- Bohr radius
- Radius of nth orbit
- Atomic number

Memory Aids:
- Radius depends on n and Z, not nuclear mass.

Quick Tips:
- Remember that isotopes have the same atomic number but different mass numbers. The radius of the orbit depends on the atomic number, not the mass number.

79 Medium

79 of 193Question permalink

Application +4 | -1

If

n = 3,\;l = 0\;\mathrm{and}\;m = 0

, then atomic number is

View answer and explanation

Correct option: D

11 or 12

The correct answer is D (11 or 12) because the quantum numbers given can correspond to either sodium (Na) or magnesium (Mg).

Solution approach

Correct Answer + Core Reason: 11 or 12. The given quantum numbers correspond to the last electron in either Na or Mg.

Step-by-Step Solution:

Step 1: Understand the quantum numbers.\nn = 3 represents the third energy level.\nl = 0 represents an s orbital.\nm = 0 is consistent with an s orbital (l=0 has only m=0).

Step 2: Determine possible electron configurations.\nThe electron configuration ends in 3s1 or 3s2.

Step 3: Identify the elements.
3s1 corresponds to Sodium (Na), which has an atomic number of 11.
3s2 corresponds to Magnesium (Mg), which has an atomic number of 12.

Step 4: Conclusion.\nThe atomic number is either 11 or 12.

Why other options are wrong:
- Option A: 12 or 13: 13 corresponds to Aluminum, which has an electron configuration ending in 3p1, not 3s1 or 3s2.
- Option B: 13 or 14: Neither 13 (Al) nor 14 (Si) have electron configurations ending in 3s1 or 3s2.
- Option C: 10 or 11: 10 corresponds to Neon, which has a full 2p orbital. 11 is correct, but 10 is not.

Key Concepts:
- Quantum numbers (n, l, m)
- Electron configuration
- Atomic number

Memory Aids:
- n = principal quantum number (energy level)
- l = azimuthal quantum number (orbital shape: 0=s, 1=p, 2=d, 3=f)
- m = magnetic quantum number (orbital orientation)

Quick Tips:
- Remember the relationship between l and the number of orbitals (2l+1).
- Know the electron configurations of common elements.

80 Medium

80 of 193Question permalink

Application +4 | -1

The relationship between the energy

{E_1}

of the radiation with a wavelength 8000Å and the energy

{E_2}

of the radiation with a wavelength 16000Å is

View answer and explanation

Correct option: B

{E_1} = 2{E_2}

Option B (

{E_1} = 2{E_2}

) is correct because energy is inversely proportional to wavelength, so doubling the wavelength halves the energy.

Solution approach

Correct Answer + Core Reason: ${E_1} = 2{E_2}$ . Energy is inversely proportional to wavelength.

Step-by-Step Solution:

Step 1: Recall the relationship between energy and wavelength: $E = hc/\lambda$

Step 2: Define $E_1$ and $E_2$ :
- $E_1 = hc/\lambda_1 = hc/8000 \,\text{\AA}$
- $E_2 = hc/\lambda_2 = hc/16000 \,\text{\AA}$

Step 3: Find the ratio $E_1/E_2$ :
- $E_1/E_2 = (hc/8000 \,\text{\AA}) / (hc/16000 \,\text{\AA}) = 16000/8000 = 2$

Step 4: Therefore, $E_1 = 2E_2$

Step 5: Final Answer: ${E_1} = 2{E_2}$

Why other options are wrong:
- Option A: ${E_1} = 6{E_2}$ : Incorrect relationship between energies.
- Option C: ${E_1} = 4{E_2}$ : Incorrect relationship between energies.
- Option D: ${E_1} = 1/2{E_2}$ : Incorrect relationship between energies. This would be true if the wavelength of E1 was twice that of E2.

Key Concepts:
- Relationship between energy and wavelength: $E = hc/\lambda$

Memory Aids:
- Remember that energy and wavelength are inversely proportional.

Quick Tips:
- When comparing energies and wavelengths, focus on the inverse relationship.

81 Medium

81 of 193Question permalink

Application +4 | -1

Pick out iso electronic species from the following
(I)

CH_3^ +

(II)

{H_3}{O^ + }

(III)

N{H_3}

(IV)

CH_3^ -

View answer and explanation

Correct option: D

II,III,IV

Option D (II,III,IV) is correct because isoelectronic species have the same number of electrons. H3O+, NH3, and CH3- all have 10 electrons.

Solution approach

Correct Answer + Core Reason: II,III,IV. Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species:
- I) $CH_3^+$ : C (6) + 3H (3) - 1 (positive charge) = 8 electrons
- II) ${H_3}{O^+}$ : 3H (3) + O (8) - 1 (positive charge) = 10 electrons
- III) $N{H_3}$ : N (7) + 3H (3) = 10 electrons
- IV) $CH_3^-$ : C (6) + 3H (3) + 1 (negative charge) = 10 electrons

Step 2: Identify the species with the same number of electrons.
- ${H_3}{O^+}$ , $N{H_3}$ , and $CH_3^-$ all have 10 electrons.

Why other options are wrong:
- Option A: I has 8 electrons, II has 10 electrons.
- Option B: I has 8 electrons, IV has 10 electrons.
- Option C: I has 8 electrons, III has 10 electrons.

Key Concepts:
- Isoelectronic species
- Counting electrons in molecules and ions

Memory Aids:
- Iso- means same.

Quick Tips:
- Account for the charge when counting electrons in ions.

82 Medium

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Understanding +4 | -1

Which particle may be removed from a stable neutral atom with least energy change?

View answer and explanation

Correct option: D

An electron

Option D (An electron) is correct because electrons are the least tightly bound particles in a neutral atom and require the least energy to remove.

Solution approach

Correct Answer + Core Reason: An electron. Electrons are the least tightly bound particles.

Step-by-Step Solution:

Step 1: Consider the structure of an atom.

Step 2: An atom consists of a nucleus (protons and neutrons) surrounded by electrons.

Step 3: Electrons are held in place by the electromagnetic force between the negatively charged electrons and the positively charged nucleus.

Step 4: Electrons are generally easier to remove than protons or neutrons because they are located outside the nucleus and are less tightly bound.

Why other options are wrong:
- Option A: An α-particle consists of 2 protons and 2 neutrons, which are tightly bound within the nucleus and require a significant amount of energy to remove.
- Option B: A neutron is located within the nucleus and requires a significant amount of energy to remove.
- Option C: A proton is located within the nucleus and requires a significant amount of energy to remove.

Key Concepts:
- Atomic structure
- Electrons, protons, and neutrons
- Ionization energy

Memory Aids:
- Electrons are easier to remove than nuclear particles.

Quick Tips:
- Remember that electrons are the outermost particles in an atom and are the easiest to remove.

83 Medium

83 of 193Question permalink

Analyzing +4 | -1

Which of the following is not permissible arrangement of electrons in an atom

View answer and explanation

Correct option: A

n = 3,\,\,l = 2,\,\,m = - 3,\,\,s = - \frac{1}{2}

Option A is the correct answer because for n=3 and l=2, the possible values of m are -2, -1, 0, 1, 2. m=-3 is not permissible.

Solution approach

Correct Answer + Core Reason: $n = 3,\,\,l = 2,\,\,m = - 3,\,\,s = - \frac{1}{2}$ . The value of m is outside the allowed range for the given l.

Step-by-Step Solution:

Step 1: Identify the given quantum numbers: $n = 3$ , $l = 2$ , $m = -3$ , $s = -\frac{1}{2}$ .

Step 2: Recall the rules for quantum numbers:
- $n$ can be any positive integer (1, 2, 3, ...).
- $l$ can range from 0 to $n-1$ .
- $m$ can range from $-l$ to $+l$ , including 0.
- $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$ .

Step 3: Check if the given quantum numbers satisfy the rules.
- For $n = 3$ , the possible values of $l$ are 0, 1, and 2. So, $l = 2$ is allowed.
- For $l = 2$ , the possible values of $m$ are -2, -1, 0, 1, and 2. The given value $m = -3$ is not allowed.
- $s = -\frac{1}{2}$ is allowed.

Step 4: Since $m = -3$ is not allowed for $l = 2$ , the given arrangement of electrons is not permissible.

Why other options are wrong:
- Option B: $n = 5,\,\,l = 3,\,\,m = 0,\,\,s = + \frac{1}{2}$ is permissible because $l$ is between 0 and $n-1$ (0 to 4), and $m$ is between $-l$ and $+l$ (-3 to 3).
- Option C: $n = 4,\,\,l = 0,\,\,m = 0,\,\,s = - \frac{1}{2}$ is permissible because $l$ is between 0 and $n-1$ (0 to 3), and $m$ is between $-l$ and $+l$ (0 to 0).
- Option D: $n = 3,\,\,l = 2,\,\,m = - 2,\,\,s = - \frac{1}{2}$ is permissible because $l$ is between 0 and $n-1$ (0 to 2), and $m$ is between $-l$ and $+l$ (-2 to 2).

Key Concepts:
- Quantum numbers (principal, azimuthal, magnetic, spin)
- Rules for allowed values of quantum numbers

Memory Aids:
- Remember the ranges for each quantum number: $n \geq 1$ , $0 \leq l \leq n-1$ , $-l \leq m \leq +l$ , $s = \pm \frac{1}{2}$

Quick Tips:
- Always check if the given values of quantum numbers fall within the allowed ranges.

84 Medium

84 of 193Question permalink

Remembering +4 | -1

For ionizing an ground state hydrogen atom, the energy needed in electron volts will be :

View answer and explanation

Correct option: A

13.6

Option A (

13.6

) is correct because the ionization energy of a ground state hydrogen atom is 13.6 eV.

Solution approach

Correct Answer + Core Reason: $13.6$ . The ionization energy of hydrogen is 13.6 eV.

Step-by-Step Solution:

Step 1: The question asks for the energy needed to ionize a ground state hydrogen atom.

Step 2: Recall that the ionization energy of hydrogen is 13.6 eV. This is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞).

Step 3: Therefore, the energy needed is 13.6 eV.

Why other options are wrong:
- Option B: $>13.6$ is incorrect because 13.6 eV is the exact energy required. More energy than this will also ionize the atom, but the question asks for the energy needed.
- Option C: $<13.6$ is incorrect because less than 13.6 eV will not ionize the atom.
- Option D: $3.4$ is incorrect because it is the energy of the electron in the n=2 state, not the ionization energy.

Key Concepts:
- Ionization energy
- Hydrogen atom energy levels

Memory Aids:
- Remember the ionization energy of hydrogen is 13.6 eV.

Quick Tips:
- Know the ionization energy of hydrogen.

85 Medium

85 of 193Question permalink

Application +4 | -1

What is the maximum number of electrons in an atom that can have the following quantum numbers

n = 4,\;{m_1} = + 1?

View answer and explanation

Correct option: D

6

Option D (6) is correct because for n=4 and ml=+1, there are 2 possible values of s (+1/2 and -1/2) for each of the three orbitals (4p, 4d) with ml=+1.

Solution approach

Correct Answer + Core Reason: 6. For a given $n$ and $m_l$ , there are a limited number of electrons possible.

Step-by-Step Solution:

Step 1: Identify the given quantum numbers: $n = 4$ and $m_l = +1$ .

Step 2: Determine the possible values of $l$ . Since $n = 4$ , $l$ can be 0, 1, 2, or 3. However, $m_l$ ranges from $-l$ to $+l$ . Since $m_l = +1$ , $l$ must be at least 1. Therefore, $l$ can be 1, 2, or 3.

Step 3: Determine the orbitals corresponding to these $l$ values:
- $l = 1$ corresponds to the $p$ orbital.
- $l = 2$ corresponds to the $d$ orbital.
- $l = 3$ corresponds to the $f$ orbital.

Step 4: For each orbital with $m_l = +1$ , there can be two electrons, one with spin $+1/2$ and one with spin $-1/2$ .

Step 5: Count the number of orbitals with $m_l = +1$ :
- For $l = 1$ ( $p$ orbital), $m_l$ can be -1, 0, +1. So, there is one orbital with $m_l = +1$ .
- For $l = 2$ ( $d$ orbital), $m_l$ can be -2, -1, 0, +1, +2. So, there is one orbital with $m_l = +1$ .
- For $l = 3$ ( $f$ orbital), $m_l$ can be -3, -2, -1, 0, +1, +2, +3. So, there is one orbital with $m_l = +1$ .

Step 6: Since each orbital with $m_l = +1$ can hold two electrons, the total number of electrons is $2 \times 3 = 6$ .

Final Answer: 6

Why other options are wrong:
- Option A (4): This is incorrect because it doesn't consider all possible orbitals with $m_l = +1$ for $n = 4$ .
- Option B (15): This is incorrect and doesn't relate to the quantum numbers given.
- Option C (3): This is incorrect because it only counts the number of orbitals with $m_l = +1$ , but not the two electrons each can hold.

Key Concepts:
- Quantum Numbers (n, l, ml, s)
- Atomic Orbitals (s, p, d, f)
- Pauli Exclusion Principle

Memory Aids:
- Remember the ranges for each quantum number: $n = 1, 2, 3,...$ ; $l = 0, 1, ..., n-1$ ; $m_l = -l, -l+1, ..., 0, ..., l-1, l$ ; $s = +1/2, -1/2$ .

Quick Tips:
- Always consider the possible values of $l$ and $m_l$ for a given $n$ .
- Remember that each orbital can hold a maximum of two electrons.

86 Medium

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Analyzing +4 | -1

Which set is not correct?

View answer and explanation

Correct option: C

3,1,2,-1/2

Option C (3,1,2,-1/2) is not correct because the value of 'l' (azimuthal quantum number) must be less than 'n' (principal quantum number), and the value of 'm' (magnetic quantum number) must be between -l and +l. Here, l=1, so m can only be -1, 0, or +1, not 2.

Solution approach

Correct Answer + Core Reason: 3,1,2,-1/2. This set of quantum numbers is not correct because the magnetic quantum number (m) cannot be greater than the azimuthal quantum number (l).

Step-by-Step Solution:

Step 1: Understand the rules for quantum numbers.\nn (principal quantum number): 1, 2, 3, ...\nl (azimuthal quantum number): 0, 1, 2, ..., n-1\nm (magnetic quantum number): -l, -l+1, ..., 0, ..., l-1, l\ns (spin quantum number): +1/2, -1/2

Step 2: Check each set of quantum numbers.\nA. 3, 1, 0, -1/2: n=3, l=1, m=0, s=-1/2. Valid.\nB. 3, 2, 1, +1/2: n=3, l=2, m=1, s=+1/2. Valid.\nC. 3, 1, 2, -1/2: n=3, l=1, m=2, s=-1/2. Invalid because |m| > l.\nD. 3, 2, 0, +1/2: n=3, l=2, m=0, s=+1/2. Valid.

Step 3: Identify the incorrect set.\nSet C is incorrect because m=2, but l=1, which violates the rule that -l ≤ m ≤ l.

Why other options are wrong:
- Option A: The set 3,1,0,-1/2 is valid as it follows all the rules for quantum numbers.
- Option B: The set 3,2,1,+1/2 is valid as it follows all the rules for quantum numbers.
- Option D: The set 3,2,0,+1/2 is valid as it follows all the rules for quantum numbers.

Key Concepts:
- Quantum numbers (n, l, m, s)
- Rules for quantum numbers

Memory Aids:
- Remember the relationships between n, l, and m: l < n and |m| ≤ l.

Quick Tips:
- Always check if the given quantum numbers follow the established rules.

87 Medium

87 of 193Question permalink

Analyzing +4 | -1

The value of charge on the oil droplets experimentally observed were

- 1.6 \times {10^{ - 19}}

and

- 4 \times {10^{ - 19}}

coulomb. The value of the electronic charge, indicated by these results is:

View answer and explanation

Correct option: D

- 0.8 \times {10^{ - 19}}

Option D (

- 0.8 \times {10^{ - 19}}

) is correct because it represents the greatest common divisor of the observed charges, suggesting it's the fundamental unit of charge.

Solution approach

Correct Answer + Core Reason: $- 0.8 \times {10^{ - 19}}$ . This is the electronic charge indicated by the results.

Step-by-Step Solution:

Step 1: Analyze the given charges: $- 1.6 \times 10^{-19} \text{{ C}}$ and $- 4 \times 10^{-19} \text{{ C}}$ .

Step 2: Find the greatest common divisor (GCD) of the magnitudes of the charges. This will give us the fundamental unit of charge.

Step 3: $1.6 \times 10^{-19} = 2 \times (0.8 \times 10^{-19})$ and $4 \times 10^{-19} = 5 \times (0.8 \times 10^{-19})$ .

Step 4: The GCD is $0.8 \times 10^{-19} \text{{ C}}$ . Since the charges are negative, the electronic charge is $- 0.8 \times 10^{-19} \text{{ C}}$ .

Why other options are wrong:
- **Option A ( $1.6 \times {10^{ - 19}}$ ): This is incorrect. While this is a multiple of the fundamental charge, it's not the smallest possible unit indicated by the data.
- Option B ( $- 2.4 \times {10^{ - 19}}$ ): This is incorrect. This value is not a common divisor of the given charges.
- Option C ( $- 4 \times {10^{ - 19}}$ ): This is incorrect. While this is a multiple of the fundamental charge, it's not the smallest possible unit indicated by the data.

Key Concepts:
- Millikan's oil drop experiment
- Quantization of charge

Memory Aids:
- The charge on any object is an integer multiple of the elementary charge.

Quick Tips:**
- Find the greatest common divisor of the observed charges to determine the elementary charge.

88 Medium

88 of 193Question permalink

Analyzing +4 | -1

Assertion: Ratio of anode rays is different for different gases.
Reason: Proton is the fundamental particle present in gases.

View answer and explanation

Correct option: B

Both assertion and reason are true but reason is not the correct explanation of assertion

Option B (Both assertion and reason are true but reason is not the correct explanation of assertion) is correct because anode rays depend on the gas used, and protons are fundamental particles, but the proton presence doesn't explain the varying ratio.

Solution approach

Correct Answer + Core Reason: Both assertion and reason are true but reason is not the correct explanation of assertion. Anode rays are gas-dependent, and protons are fundamental, but the latter doesn't explain the former.

Step-by-Step Solution:

89 Medium

89 of 193Question permalink

Analyzing +4 | -1

The last electron placed in the third (n=3) quantum shell for:

View answer and explanation

Correct option: C

Cu

Option C (Cu) is correct because copper has an electronic configuration of [Ar] 3d¹⁰4s¹, meaning the last electron enters the n=3 shell (3d orbital).

Solution approach

Correct Answer + Core Reason: Cu. Copper has its last electron placed in the third quantum shell (n=3).

Step-by-Step Solution:

Step 1: Write the electronic configurations for each element.\nKr: [Ar] 4s² 3d¹⁰ 4p⁶ (last electron in n=4)\nZn: [Ar] 4s² 3d¹⁰ (last electron in n=4, but also in n=3)\nCu: [Ar] 4s¹ 3d¹⁰ (last electron in n=3)\nCa: [Ar] 4s² (last electron in n=4)

Step 2: Identify the element with the last electron in the n=3 shell.\nCopper has its last electron in the 3d orbital, which is in the n=3 shell.

Why other options are wrong:
- Option A: The last electron in Kr is in the 4p orbital (n=4).
- Option B: The last electron in Zn is in the 4s orbital (n=4), although it also has electrons in 3d.
- Option D: The last electron in Ca is in the 4s orbital (n=4).

Key Concepts:
- Electronic configuration
- Quantum shells (n)

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

Quick Tips:
- Write out the electronic configurations to easily determine the quantum shell of the last electron.

90 Medium

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Understanding +4 | -1

When atoms are bombarded with α-particles suffer deflections while others pass through undeflected.
This is because :

View answer and explanation

Correct option: B

The nucleus occupies much smaller volume compared to the volume of the atom

Option B (The nucleus occupies much smaller volume compared to the volume of the atom) is correct because most of the atom is empty space, allowing most alpha particles to pass through undeflected, while the small, dense nucleus causes deflections.

Solution approach

Correct Answer + Core Reason: The nucleus occupies much smaller volume compared to the volume of the atom. This explains why most alpha particles pass through undeflected.

Step-by-Step Solution:

Step 1: Describe Rutherford's gold foil experiment.
- Alpha particles were directed at a thin gold foil.
- Most alpha particles passed through undeflected.
- A small fraction were deflected at large angles.

Step 2: Explain the implications of the results.
- Most of the atom is empty space, allowing most alpha particles to pass through undeflected.
- The positive charge and most of the mass are concentrated in a small, dense nucleus.
- Alpha particles that pass close to the nucleus are deflected due to electrostatic repulsion.

Why other options are wrong:
- Option A: The force of attraction on the α-particle by the oppositely charged electrons is not sufficient is incorrect because the electrons are distributed throughout the atom and their effect is minimal compared to the concentrated positive charge of the nucleus.
- Option C: The force of repulsion on the fast moving α-particle small is incorrect because the force of repulsion is significant when the alpha particle gets close enough to the nucleus.
- Option D: The effect in the nucleus do not have any effect on the α-particles is incorrect because the nucleus is responsible for the deflection of alpha particles.

Key Concepts:
- Rutherford's gold foil experiment
- Atomic structure (nucleus and electrons)
- Electrostatic repulsion

Memory Aids:
- Remember that most of the atom is empty space.

Quick Tips:
- The large-angle deflections in Rutherford's experiment provided evidence for the existence of the nucleus.

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