Atoms of different elements having same mass number

Atoms of same elements having same mass number

Atoms of different elements having same number of neutrons

STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Easy

1 Easy

1 of 176Question permalink

Understanding +4 | -1

Which of the following pairs have identical values of e/m?

View answer and explanation

Correct option: A

Deuterium and an α-particles

Option A (Deuterium and an α-particles) is correct because both have a charge-to-mass ratio of approximately 1/2.

Solution approach

Correct Answer + Core Reason: Deuterium and an α-particles. Both have the same e/m ratio.

Step-by-Step Solution:

Step 1: Define the e/m ratio (specific charge).

Step 2: Calculate the e/m ratio for each particle:
- Proton: e/m = e/mp
- Deuterium: e/m = e/(2mp) (Deuterium has 1 proton and 1 neutron)
- Alpha particle: e/m = 2e/(4mp) = e/(2mp) (Alpha particle has 2 protons and 2 neutrons)
- Neutron: e/m = 0/mn = 0
- Electron: e/m = e/me
- Gamma rays: Gamma rays are photons and have no mass or charge.

Step 3: Compare the e/m ratios.
- Deuterium and alpha particles have the same e/m ratio: e/(2mp).

Why other options are wrong:
- Option B: A proton and deuterium have different e/m ratios. Proton has e/mp and deuterium has e/(2mp).
- Option C: A proton and a neutron have different e/m ratios. Proton has e/mp and neutron has 0.
- Option D: An electron and $\gamma -$ rays have different e/m characteristics. Electron has e/me and gamma rays have no mass or charge.

Key Concepts:
- Specific charge (e/m ratio)
- Composition of subatomic particles (protons, neutrons, electrons, alpha particles, deuterium)

Memory Aids:
- Remember the composition of deuterium (1 proton, 1 neutron) and alpha particles (2 protons, 2 neutrons).

Quick Tips:
- The e/m ratio depends on both the charge and the mass of the particle.

2 Easy

2 of 176Question permalink

Remembering +4 | -1

The correct de Broglie relationship is:

View answer and explanation

Correct option: B

\lambda = \frac{h}{{mu}}

Option B (

\lambda = \frac{h}{{mu}}

) is correct because it represents the correct de Broglie relationship between wavelength, Planck's constant, mass, and velocity.

Solution approach

Correct Answer + Core Reason: $\lambda = \frac{h}{{mu}}$ . This is the correct de Broglie relationship.

Step-by-Step Solution:

Step 1: Recall the de Broglie equation.

Step 2: The de Broglie equation is given by $\lambda = \frac{h}{p}$ , where $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum.

Step 3: Momentum $p = mu$ , where $m$ is the mass and $u$ is the velocity.

Step 4: Substitute $p = mu$ into the de Broglie equation: $\lambda = \frac{h}{mu}$ .

Why other options are wrong:
- Option A: $\frac{\lambda }{{mu}} = p$ : This is incorrect. It's a rearrangement of the de Broglie equation, but not in the standard form.
- Option C: $\lambda = \frac{h}{{mp}}$ : This is incorrect. It uses momentum in the denominator twice.
- Option D: $\lambda m = \frac{u}{p}$ : This is incorrect. It's not a valid form of the de Broglie equation.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- $\lambda = \frac{h}{mu}$ (de Broglie equation)

Quick Tips:
- Remember the standard form of the de Broglie equation.

3 Easy

3 of 176Question permalink

Understanding +4 | -1

From the discharge tube experiment, it is concluded that:

View answer and explanation

Correct option: B

Matter contains electrons

Option B (Matter contains electrons) is correct because the discharge tube experiment led to the discovery of electrons, demonstrating that matter contains these negatively charged particles.

Solution approach

Correct Answer + Core Reason: Matter contains electrons. The discharge tube experiment led to the discovery of electrons.

Step-by-Step Solution:

Step 1: Recall the discharge tube experiment.

Step 2: The discharge tube experiment involved passing electricity through a gas at low pressure.

Step 3: This experiment led to the discovery of cathode rays, which were later identified as electrons.

Step 4: The discovery of electrons showed that matter contains electrons.

Why other options are wrong:
- Option A: The mass of a proton is not in fraction; it's approximately 1 atomic mass unit.
- Option C: The discharge tube experiment did not directly prove that the nucleus contains a positive charge (Rutherford's experiment did).
- Option D: Positive rays are heavier than protons is not a direct conclusion from the discharge tube experiment.

Key Concepts:
- Discharge tube experiment
- Cathode rays
- Electrons

Memory Aids:
- The discharge tube experiment led to the discovery of electrons.

Quick Tips:
- Remember the key findings of the discharge tube experiment.

4 Easy

4 of 176Question permalink

Analyzing +4 | -1

The compound in which cation is isoelectronic with anion is

View answer and explanation

Correct option: D

{K_2}S

Option D (

{K_2}S

) is correct because both

K^+

and

S^{2-}

have the same number of electrons as Argon, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: ${K_2}S$ . Both $K^+$ and $S^{2-}$ are isoelectronic with Argon.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each ion for each compound.

Step 2: For NaCl: $Na^+$ has 10 electrons, $Cl^-$ has 18 electrons. Not isoelectronic.

Step 3: For CsF: $Cs^+$ has 54 electrons, $F^-$ has 10 electrons. Not isoelectronic.

Step 4: For NaI: $Na^+$ has 10 electrons, $I^-$ has 54 electrons. Not isoelectronic.

Step 5: For ${K_2}S$ : $K^+$ has 18 electrons, $S^{2-}$ has 18 electrons. Isoelectronic.

Step 6: Conclusion: Only ${K_2}S$ has cation and anion that are isoelectronic.

Why other options are wrong:
- Option A: $Na^+$ and $Cl^-$ have different number of electrons (10 and 18 respectively).
- Option B: $Cs^+$ and $F^-$ have different number of electrons (54 and 10 respectively).
- Option C: $Na^+$ and $I^-$ have different number of electrons (10 and 54 respectively).

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

Memory Aids:
- Cations lose electrons, anions gain electrons.

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

5 Easy

5 of 176Question permalink

Remembering +4 | -1

The electron possesses wave properties was shown experimentally by:

View answer and explanation

Correct option: C

Davission and germer

Option C is correct because Davisson and Germer experimentally demonstrated the wave nature of electrons through electron diffraction.

Solution approach

Correct Answer + Core Reason: Davission and germer. Davisson and Germer experimentally showed that electrons possess wave properties.

Step-by-Step Solution:

Step 1: Recall the historical experiments related to the wave nature of electrons.
- de Broglie proposed the wave-particle duality of matter, suggesting that electrons have wave-like properties.
- Davisson and Germer performed an experiment where they observed the diffraction of electrons by a nickel crystal, providing experimental evidence for de Broglie's hypothesis.
- Bohr developed a model of the atom based on quantized energy levels.
- Schrödinger developed the wave equation to describe the behavior of electrons in atoms.

Step 2: Identify the experiment that demonstrated the wave properties of electrons.
- The Davisson-Germer experiment provided experimental evidence for the wave nature of electrons through electron diffraction.

Why other options are wrong:
- Option A: Bohr proposed a model of the atom, but did not experimentally demonstrate the wave nature of electrons.
- Option B: de Broglie proposed the wave-particle duality, but did not experimentally verify it.
- Option D: Schrödinger developed the wave equation, but did not experimentally demonstrate the wave nature of electrons.

Key Concepts:
- Wave-particle duality
- Electron diffraction
- Davisson-Germer experiment

Memory Aids:
- Remember that Davisson and Germer's experiment provided experimental proof of de Broglie's hypothesis.

Quick Tips:
- Understand the historical context of the development of quantum mechanics.

6 Easy

6 of 176Question permalink

Analyzing +4 | -1

Kinetic energy of an electron in hydrogen atom increases after transition from an orbit

n_1

to another orbit

n_2

. Then

View answer and explanation

Correct option: B

n_l>n_2

Option B (

n_l>n_2

) is correct because the kinetic energy of an electron increases as it transitions to a lower energy level (closer to the nucleus).

Solution approach

Correct Answer + Core Reason: $n_l>n_2$ . Kinetic energy increases when the electron moves closer to the nucleus (lower energy level).

Step-by-Step Solution:

Step 1: Recall that the kinetic energy (KE) of an electron in a hydrogen atom is related to its potential energy (PE) by $KE = -\frac{1}{2} PE$ . Also, the total energy $E = KE + PE = \frac{1}{2} PE = -KE$ .

Step 2: The total energy of an electron in the nth orbit is given by $E_n \propto -\frac{1}{n^2}$ . Therefore, as n decreases, the total energy becomes more negative (lower energy).

Step 3: Since $E = -KE$ , as the total energy becomes more negative, the kinetic energy becomes more positive (increases).

Step 4: If the kinetic energy increases after the transition from orbit $n_1$ to $n_2$ , it means the electron has moved to a lower energy level, which implies $n_2 < n_1$ .

Why other options are wrong:
- Option A: This is incorrect because if $n_1 < n_2$ , the electron would be moving to a higher energy level, and its kinetic energy would decrease.
- Option C: This is incorrect because we can predict the relationship between $n_1$ and $n_2$ based on the change in kinetic energy.
- Option D: This is incorrect because there is a definite relationship between the change in kinetic energy and the energy levels.

Key Concepts:
- Bohr's model of the atom
- Energy levels in hydrogen atom
- Kinetic energy and potential energy of an electron

Memory Aids:
- Remember that as the electron moves closer to the nucleus, its kinetic energy increases, and its potential energy becomes more negative.

Quick Tips:
- Always remember the relationship between kinetic energy, potential energy, and total energy in the Bohr model.

7 Easy

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Understanding +4 | -1

The Heisenberg^' s uncertainty principle can be applied to:

View answer and explanation

Correct option: D

An electron

Option D (An electron) is correct because Heisenberg's uncertainty principle is most significant for particles with very small mass, like electrons.

Solution approach

Correct Answer + Core Reason: An electron. Heisenberg's uncertainty principle is significant for particles with small mass.

Step-by-Step Solution:

Step 1: The question asks to which of the given objects Heisenberg's uncertainty principle can be applied.

Step 2: Recall that Heisenberg's uncertainty principle is significant for particles with very small mass, such as electrons.

Step 3: Consider the masses of the given objects:
- Cricket ball: Relatively large mass.
- Football: Relatively large mass.
- Jet aeroplane: Very large mass.
- Electron: Very small mass.

Step 4: Identify the object with the smallest mass.

Step 5: The electron is the correct answer.

Why other options are wrong:
- Option A (A cricket ball): This is incorrect. The uncertainty principle is negligible for macroscopic objects like cricket balls.
- Option B (A football): This is incorrect. The uncertainty principle is negligible for macroscopic objects like footballs.
- Option C (A jet aeroplane): This is incorrect. The uncertainty principle is negligible for macroscopic objects like jet aeroplanes.

Key Concepts:
- Heisenberg's uncertainty principle
- Quantum mechanics
- Mass and uncertainty

Memory Aids:
- Remember that the uncertainty principle is most important for very small particles.

Quick Tips:
- The uncertainty principle is generally not noticeable for everyday objects.

8 Easy

8 of 176Question permalink

Remembering +4 | -1

Positive charge in an atom is:

View answer and explanation

Correct option: B

Concentrated in the nucleus

Option B is correct because the positive charge in an atom is concentrated in the nucleus due to the presence of protons.

Solution approach

Correct Answer + Core Reason: Concentrated in the nucleus. The positive charge in an atom is due to protons, which are located in the nucleus.

Step-by-Step Solution:

Step 1: Recall the structure of an atom.

Step 2: Identify the location of protons in an atom.

Step 3: Protons are located in the nucleus, so the positive charge is concentrated in the nucleus.

Why other options are wrong:
- Option A: Positive charge is not scattered all over the atom.
- Option C: Electrons revolve around the nucleus, not positive charges.
- Option D: The correct answer is true.

Key Concepts:
- Atomic structure
- Nucleus
- Protons

Memory Aids:
- Protons are positive and in the nucleus.

Quick Tips:
- Remember the location of subatomic particles.

9 Easy

9 of 176Question permalink

Understanding +4 | -1

The nodes present in 3p-orbitals are

View answer and explanation

Correct option: A

One spherical, one planar

Option A (One spherical, one planar) is correct because a 3p-orbital has one radial node (spherical) and one angular node (planar).

Solution approach

Correct Answer + Core Reason: One spherical, one planar. 3p-orbitals have one spherical and one planar node.

Step-by-Step Solution:

Step 1: Calculate the number of radial nodes: Radial nodes = n - l - 1. For 3p, n = 3 and l = 1. Therefore, radial nodes = 3 - 1 - 1 = 1 (spherical node).

Step 2: Calculate the number of angular nodes: Angular nodes = l. For 3p, l = 1. Therefore, angular nodes = 1 (planar node).

Step 3: Final Answer: The 3p-orbital has one spherical node and one planar node.

Why other options are wrong:
- Option B: This option is incorrect because it states there are two spherical nodes.
- Option C: This option is incorrect because it states there are two planar nodes.
- Option D: This option is incorrect because it only mentions one planar node and doesn't account for the spherical node.

Key Concepts:
- Radial nodes
- Angular nodes
- Principal quantum number (n)
- Azimuthal quantum number (l)

Memory Aids:
- Radial nodes = n - l - 1
- Angular nodes = l

Quick Tips:
- Remember the formulas for calculating radial and angular nodes.

10 Easy

10 of 176Question permalink

Understanding +4 | -1

Visible spectrum of hydrogen shows that it exists in two different forms which are based on direction of spin of the:

View answer and explanation

Correct option: B

Nuclei of hydrogen atoms

Option B (Nuclei of hydrogen atoms) is correct because the two forms of hydrogen are based on the spin of the nuclei (ortho and para hydrogen).

Solution approach

Correct Answer + Core Reason: Nuclei of hydrogen atoms. The two forms of hydrogen are based on the spin of the nuclei.

Step-by-Step Solution:

Step 1: The question asks about the basis for the two different forms of hydrogen in the visible spectrum.

Step 2: Recall that hydrogen exists in two forms, ortho-hydrogen and para-hydrogen, which are based on the spin of the nuclei of the hydrogen atoms.

Step 3: In ortho-hydrogen, the spins of the two nuclei are parallel, while in para-hydrogen, the spins are anti-parallel.

Why other options are wrong:
- Option A: Molecule of hydrogen is incorrect because the forms are based on the nuclei, not the entire molecule.
- Option C: Electrons of hydrogen is incorrect because the forms are based on the nuclei, not the electrons.
- Option D: Atoms of hydrogen molecule is incorrect because the forms are based on the nuclei, not the entire atoms.

Key Concepts:
- Ortho-hydrogen
- Para-hydrogen
- Nuclear spin

Memory Aids:
- Ortho: same spin
- Para: opposite spin

Quick Tips:
- Remember the difference between ortho and para hydrogen.

11 Easy

11 of 176Question permalink

Analyzing +4 | -1

Assertion :

{ }_{11}^{22} {Na}

emits a position giving

{ }_{12}^{22} {Mg}

.

Reason : In

\beta^{+}

emission neutron is transformed into proton.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D (If the assertion and reason both are false) is correct because in

\beta^+

emission, a proton is transformed into a neutron, and the atomic number decreases, not increases as suggested by the assertion.

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The assertion is false because Na-22 emits a positron to become Ne-22, and the reason is false because in beta+ decay, a proton converts to a neutron.

Step-by-Step Solution:

Step 1: Analyze the assertion:
- ${ }_{11}^{22} {Na}$ emitting a positron should result in a decrease in atomic number (number of protons) by 1, not an increase. The correct product should be ${ }_{10}^{22} {Ne}$ , not ${ }_{12}^{22} {Mg}$ . Therefore, the assertion is false.

Step 2: Analyze the reason:
- In $\beta^+$ emission (positron emission), a proton is transformed into a neutron, a positron, and a neutrino. The correct transformation is $p \rightarrow n + e^+ + \nu$ . Therefore, the reason is false.

Step 3: Conclusion: Both the assertion and the reason are false.

Step 4: Final Answer: If the assertion and reason both are false.

Why other options are wrong:
- Option A: If both assertion and reason are true and the reason is the correct explanation of the assertion: Both are false.
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: Both are false.
- Option C: If assertion is true but reason is false: The assertion is also false.

Key Concepts:
- Nuclear reactions
- Positron emission ( $\beta^+$ decay)
- Atomic number and mass number

Memory Aids:
- Remember that in $\beta^+$ decay, the atomic number decreases by 1.

Quick Tips:
- Always check the conservation of mass number and atomic number in nuclear reactions.

12 Easy

12 of 176Question permalink

Application +4 | -1

The number of s-electrons present in an ion with 55
protons in its unipositive state is

View answer and explanation

Correct option: A

10

Option A (10) is correct because an element with 55 protons in its unipositive state has 54 electrons, and the number of s-electrons can be determined from the electronic configuration.

Solution approach

Correct Answer + Core Reason: 10. The electronic configuration determines the number of s-electrons.

Step-by-Step Solution:

Step 1: The element with 55 protons is Cesium (Cs). In its neutral state, it has 55 electrons.

Step 2: In its unipositive state (Cs+), it has 54 electrons.

Step 3: Write the electronic configuration of Cs+: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶

Step 4: Count the number of s-electrons: 1s² (2) + 2s² (2) + 3s² (2) + 4s² (2) + 5s² (2) = 10

Why other options are wrong:
- Option B: Incorrect count of s-electrons.
- Option C: Incorrect count of s-electrons.
- Option D: Incorrect count of s-electrons.

Key Concepts:
- Electronic configuration
- s-electrons
- Ions

Memory Aids:
- Remember the order of filling orbitals.

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

13 Easy

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Application +4 | -1

If threshold wavelength

\left( {{\lambda ^0}} \right)

for ejection of electron from metal is 330 nm. Then work function for the photo electric emission is

View answer and explanation

Correct option: C

6 \times {10^{ - 19}}J

Option C is correct. The work function is calculated using the formula φ = hc/λ, resulting in approximately 6 × 10^-19 J.

Solution approach

Correct Answer + Core Reason: $6 \times {10^{ - 19}}J$ . The work function is calculated using the formula $\phi = \frac{hc}{\lambda}$ .

Step-by-Step Solution:

Step 1: Recall the formula for work function: $\phi = \frac{hc}{\lambda}$ , where $\phi$ is the work function, $h = 6.626 \times 10^{-34} \text{ Js}$ , $c = 3 \times 10^8 \text{ m/s}$ , and $\lambda = 330 \text{ nm} = 330 \times 10^{-9} \text{ m}$ .

Step 2: Substitute the values: $\phi = \frac{(6.626 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{330 \times 10^{-9} \text{ m}} = \frac{19.878 \times 10^{-26}}{330 \times 10^{-9}} \text{ J} = 6.024 \times 10^{-19} \text{ J}$ .

Step 3: Round the result: $\phi \approx 6 \times 10^{-19} \text{ J}$ .

Why other options are wrong:
- Option A: $1.2 \times {10^{ - 18}}J$ : Incorrect; this value is too high and likely a calculation error.
- Option B: $1.2 \times {10^{ - 20}}J$ : Incorrect; this value is too low and likely a calculation error.
- Option D: $6 \times {10^{ - 12}}J$ : Incorrect; this value is too high and likely a calculation error.

Key Concepts:
- Work function
- Threshold wavelength
- Planck's constant
- Speed of light

Memory Aids:
- Remember the formula $\phi = \frac{hc}{\lambda}$ .

Quick Tips:
- Ensure units are consistent before calculating. Use the correct values for Planck's constant and the speed of light.

14 Easy

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Analyzing +4 | -1

If

{E_e}\,,\,{E_\alpha }\,\& {E_P}

,

{E_\alpha }

and

{E_P}

represents the kinetic energies of electron, -particle and a proton de Broglie wavelength then

View answer and explanation

Correct option: D

{E_e}\,\, > {E_P}\,\, > = {E_\alpha }

Option D (

{E_e}\,\, > {E_P}\,\, > = {E_\alpha }

) is correct because the kinetic energy is inversely proportional to the mass when de Broglie wavelength is constant.

Solution approach

Correct Answer + Core Reason: ${E_e}\,\, > {E_P}\,\, > = {E_\alpha }$ . Kinetic energy is inversely proportional to mass for the same de Broglie wavelength.

Step-by-Step Solution:

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

Step 2: Rearrange the equation to express kinetic energy in terms of wavelength and mass: $E = \frac{h^2}{2m\lambda^2}$ .

Step 3: Since the de Broglie wavelength is the same for all particles, $E \propto \frac{1}{m}$ .

Step 4: Compare the masses of electron, proton, and alpha particle: $m_e < m_p < m_\alpha$ .

Step 5: Therefore, the kinetic energies are related as: $E_e > E_p > E_\alpha$ .

Why other options are wrong:
- Option A: ${E_e}\,\, = {E_\alpha }\,\, = {E_P}$ : This is incorrect because the kinetic energies are inversely proportional to the masses, and the masses are different.
- Option B: ${E_e}\,\, > {E_\alpha }\,\, > = {E_P}$ : This is incorrect because the proton's kinetic energy should be greater than the alpha particle's kinetic energy.
- Option C: ${E_\alpha }\,\, > {E_{_P}}\,\, > {E_e}$ : This is incorrect because the electron has the smallest mass and thus the largest kinetic energy.

Key Concepts:
- De Broglie wavelength
- Kinetic energy
- Mass-energy relationship

Memory Aids:
- Remember that lighter particles have higher kinetic energy for the same wavelength.

Quick Tips:
- When comparing kinetic energies with the same de Broglie wavelength, focus on the inverse relationship with mass.

15 Easy

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Application +4 | -1

The uncertainty in the velocity of a particle of mass

6.626\,\,x\,\,{10^{^{ - 28}}}Kg\,\,is\,\,1\,\,x\,\,{10^{^{ - 6}}}

m/s. What is the uncertainty in its position ( in nm )

View answer and explanation

Correct option: D

1/4\pi

Option D (

1/4\pi

) is correct because Heisenberg's uncertainty principle states that

\Delta x \Delta p \geq \frac{h}{{4\pi}}

, where

\Delta p = m \Delta v

. Solving for

\Delta x

gives

\Delta x = \frac{h}{{4\pi m \Delta v}} = \frac{{6.626 \times 10^{-34}}}{{4\pi \times 6.626 \times 10^{-28} \times 1 \times 10^{-6}}} = \frac{1}{{4\pi}} \times 10^{-9} m = \frac{1}{{4\pi}} nm

.

Solution approach

Correct Answer + Core Reason: $1/4\pi$ . The uncertainty in position is calculated using Heisenberg's uncertainty principle.

Step-by-Step Solution:

Step 1: Recall Heisenberg's uncertainty principle: $\Delta x \Delta p \geq \frac{h}{{4\pi}}$ , where $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum.

Step 2: Express the uncertainty in momentum as $\Delta p = m \Delta v$ , where m is the mass and $\Delta v$ is the uncertainty in velocity.

Step 3: Substitute $\Delta p = m \Delta v$ into the uncertainty principle: $\Delta x (m \Delta v) \geq \frac{h}{{4\pi}}$

Step 4: Solve for $\Delta x$ : $\Delta x \geq \frac{h}{{4\pi m \Delta v}}$

Step 5: Substitute the given values: $h = 6.626 \times 10^{-34} \text{{ Js}}$ , $m = 6.626 \times 10^{-28} \text{{ kg}}$ , $\Delta v = 1 \times 10^{-6} \text{{ m/s}}$
Δ x ≥ \frac{{6.626 × 10^-34}}{{4π (6.626 × 10^-28)(1 × 10^-6)}} $\n\n**Step 6:** Simplify the expression:$ Δ x ≥ $\frac{1}{{4π}}$ × 10^-9 \text{{ m}} $\n\n**Step 7:** Convert to nanometers:$ Δ x ≥ $\frac{1}{{4π}}$ \text{{ nm}} $\n\n**Why other options are wrong:**\n- **Option A:**$ 1/2π $: This is incorrect because it doesn't include the factor of 4 in the denominator from the uncertainty principle.\n- **Option B:**$ 2.5/π $: This is incorrect and doesn't follow from the correct application of the uncertainty principle.\n- **Option C:**$ 5/π $: This is incorrect and doesn't follow from the correct application of the uncertainty principle.\n\n**Key Concepts:**\n- Heisenberg's uncertainty principle\n- Uncertainty in position\n- Uncertainty in momentum\n\n**Memory Aids:**\n- Remember the uncertainty principle:$ Δ x Δ p ≥ $\frac{h}{{4π}}$ $\n\n**Quick Tips:**\n- Always use the correct units and remember to convert to the desired units at the end.$

16 Easy

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Analyzing +4 | -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Loss of electron from hydrogen atom results in nucleus of

\sim 1.5 \times 10^{-3} \, \mathrm{pm}

size.

Reason R: Proton

\left(\mathrm{H}^{+}\right)

always exists in combined form.

In the light of the above statements, choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: D

Both A and R are correct but R is NOT the
correct explanation of A

Option D (Both A and R are correct but R is NOT the correct explanation of A) is correct because both the assertion and the reason are true statements, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: Both A and R are correct but R is NOT the correct explanation of A. Assertion and Reason are both correct, but unrelated.

Step-by-Step Solution:

Step 1: Analyze Assertion A: Loss of an electron from a hydrogen atom results in a proton. The size of a proton is approximately $1.5 \times 10^{-3} \, \mathrm{pm}$ . This statement is correct.

Step 2: Analyze Reason R: Protons ( $\mathrm{H}^+$ ) always exist in combined form. This statement is also correct. Protons are highly reactive and typically found bonded to other atoms or molecules, especially in aqueous solutions where they form hydronium ions ( $\mathrm{H}_3\mathrm{O}^+$ ).

Step 3: Determine if Reason R explains Assertion A: The fact that protons exist in combined form does not explain why the size of the nucleus (proton) is $1.5 \times 10^{-3} \, \mathrm{pm}$ . The size is a fundamental property of the proton itself.

Why other options are wrong:
- Option A: This is incorrect because the reason does not explain the assertion.
- Option B: This is incorrect because both the assertion and the reason are correct.
- Option C: This is incorrect because both the assertion and the reason are correct.

Key Concepts:
- Structure of the atom
- Properties of protons
- Existence of protons in combined form

Memory Aids:
- Remember that protons are fundamental particles and their size is an intrinsic property.

Quick Tips:
- Carefully analyze both the assertion and the reason to determine if they are individually correct and if the reason provides a valid explanation for the assertion.

17 Easy

17 of 176Question permalink

Analyzing +4 | -1

Which among the following species have the same number of electrons in its outermost as well as penultimate shell?

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{a}^{2 + }}

Option D (

\mathrm{C}{\mathrm{a}^{2 + }}

) is correct because Calcium (Ca) loses two electrons to form Ca2+, resulting in the same number of electrons (8) in both the outermost (2p) and penultimate (3p) shells.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{a}^{2 + }}$ . Calcium (Ca) loses 2 electrons to form Ca2+, resulting in the same number of electrons in the outermost and penultimate shells.

Step-by-Step Solution:

Step 1: Write the electronic configuration of each species.\nMg2+ (Z=12-2=10): $1s^2 2s^2 2p^6$ \nO2- (Z=8+2=10): $1s^2 2s^2 2p^6$ \nF- (Z=9+1=10): $1s^2 2s^2 2p^6$ \nCa2+ (Z=20-2=18): $1s^2 2s^2 2p^6 3s^2 3p^6$

Step 2: Identify the number of electrons in the outermost and penultimate shells.
- For Mg2+, the outermost shell is the 2nd shell (n=2), which has 8 electrons (2s2 2p6). The penultimate shell is the 1st shell (n=1), which has 2 electrons (1s2). They are not the same.
- For O2-, the outermost shell is the 2nd shell (n=2), which has 8 electrons (2s2 2p6). The penultimate shell is the 1st shell (n=1), which has 2 electrons (1s2). They are not the same.
- For F-, the outermost shell is the 2nd shell (n=2), which has 8 electrons (2s2 2p6). The penultimate shell is the 1st shell (n=1), which has 2 electrons (1s2). They are not the same.
- For Ca2+, the outermost shell is the 3rd shell (n=3), which has 8 electrons (3s2 3p6). The penultimate shell is the 2nd shell (n=2), which has 8 electrons (2s2 2p6). They are the same.

Why other options are wrong:
- Option A: $\mathrm{M}{\mathrm{g}^{2 + }}$ : The number of electrons in the outermost and penultimate shells are not the same.
- Option B: ${\mathrm{O}^{2 - }}$ : The number of electrons in the outermost and penultimate shells are not the same.
- Option C: ${\mathrm{F}^ - }$ : The number of electrons in the outermost and penultimate shells are not the same.

Key Concepts:
- Electronic configuration
- Outermost and penultimate shells

Memory Aids:
- Remember the electronic configuration rules.

Quick Tips:
- Write the electronic configuration first.
- Identify the outermost and penultimate shells.

18 Easy

18 of 176Question permalink

Application +4 | -1

Suppose

{10^{ - 17}}

J of light energy is needed by the interior of a human eye to see an object. Calculate the number of photons of green light (

({\lambda _\;} = 550

nm) needed to generate this minimum amount of energy

View answer and explanation

Correct option: C

28

Option C is correct. The number of photons is calculated by dividing the total energy needed by the energy of a single photon, resulting in approximately 28 photons.

Solution approach

Correct Answer + Core Reason: 28. The number of photons is calculated by dividing the total energy by the energy per photon.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon: $E = \frac{hc}{\lambda}$ , where $h = 6.626 \times 10^{-34} \text{ Js}$ , $c = 3 \times 10^8 \text{ m/s}$ , and $\lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m}$ .

Step 2: Substitute the values: $E = \frac{(6.626 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{550 \times 10^{-9} \text{ m}} = \frac{19.878 \times 10^{-26}}{550 \times 10^{-9}} \text{ J} = 3.614 \times 10^{-19} \text{ J}$ .

Step 3: Calculate the number of photons: $N = \frac{\text{Total energy}}{\text{Energy per photon}} = \frac{10^{-17} \text{ J}}{3.614 \times 10^{-19} \text{ J}} = 27.67 \approx 28$ .

Why other options are wrong:
- Option A: 26: Incorrect; the calculation yields approximately 28 photons.
- Option B: 27: Incorrect; the calculation yields approximately 28 photons.
- Option D: 29: Incorrect; the calculation yields approximately 28 photons.

Key Concepts:
- Energy of a photon
- Wavelength
- Number of photons

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

Quick Tips:
- Ensure units are consistent before calculating. Round to the nearest whole number for the number of photons.

19 Easy

19 of 176Question permalink

Remembering +4 | -1

Isotopes are

View answer and explanation

Correct option: C

Atoms of same elements having different mass number

Option C is correct because isotopes are atoms of the same element that have different mass numbers due to differing numbers of neutrons.

Solution approach

Correct Answer + Core Reason: Atoms of same elements having different mass number. Isotopes are atoms of the same element with the same atomic number but different mass numbers due to different numbers of neutrons.

Step-by-Step Solution:

Step 1: Define isotopes.

Step 2: Relate isotopes to atomic number and mass number.

Step 3: Isotopes have the same atomic number but different mass numbers.

Why other options are wrong:
- Option A: Isotopes are atoms of the same element, not different elements.
- Option B: Isotopes have different mass numbers, not the same mass number.
- Option D: Isotopes have the same number of protons, not the same number of neutrons (although the number of protons is the same for all isotopes of an element).

Key Concepts:
- Isotopes
- Atomic number
- Mass number

Memory Aids:
- Isotopes have the same number of protons but different numbers of neutrons.

Quick Tips:
- Remember the definition of isotopes.

20 Easy

20 of 176Question permalink

Understanding +4 | -1

As the number of orbit increase from the nucleus, the difference between the adjacent energy levels:

View answer and explanation

Correct option: C

Decreases

Option C (Decreases) is correct because as the number of orbits increases, the energy levels get closer together, thus the difference between adjacent energy levels decreases.

Solution approach

Correct Answer + Core Reason: Decreases. The difference between adjacent energy levels decreases as the number of orbits increases.

Step-by-Step Solution:

Step 1: The question asks about the change in the difference between adjacent energy levels as the number of orbits increases.

Step 2: Recall the energy levels in the hydrogen atom: $E_n = -\frac{13.6 \text{{ eV}}}{{n^2}}$ , where $n$ is the principal quantum number.

Step 3: As $n$ increases, the energy levels get closer together. This means the difference between adjacent energy levels decreases.

Step 4: Therefore, the difference between adjacent energy levels decreases.

Why other options are wrong:
- Option A (Increases): This is incorrect. The energy levels get closer together, so the difference decreases.
- Option B (Remains constant): This is incorrect. The energy levels do not remain equally spaced.
- Option D (None of these): This is incorrect. The difference decreases.

Key Concepts:
- Energy levels
- Bohr's model
- Quantum mechanics

Memory Aids:
- Remember that energy levels converge as $n$ increases.

Quick Tips:
- The energy levels are not equally spaced; they get closer together at higher energy levels.

21 Easy

21 of 176Question permalink

Remembering +4 | -1

Cathode-ray tube is used in:

View answer and explanation

Correct option: C

A television set

Option C (A television set) is correct because cathode-ray tubes (CRTs) were historically used as the display component in television sets.

Solution approach

Correct Answer + Core Reason: A television set. Cathode-ray tubes were commonly used in television sets to display images.

Step-by-Step Solution:

Step 1: Cathode-Ray Tube (CRT) Function: A CRT is a vacuum tube that produces images by directing a beam of electrons onto a phosphorescent screen.

Step 2: CRT in Television Sets: In a television set, the CRT's electron beam scans across the screen, and the intensity of the beam is modulated to create the image. The screen is coated with phosphors that emit light when struck by the electrons.

Why other options are wrong:
- Option A: Compound microscope. Compound microscopes use lenses to magnify images, not cathode-ray tubes.
- Option B: A radio receiver. Radio receivers use electronic circuits to detect and amplify radio signals, not cathode-ray tubes for display.
- Option D: A van de Graff generator. A Van de Graaff generator is used to produce high voltages, not to display images.

Key Concepts:
- Cathode-ray tube (CRT)
- Television technology

Memory Aids:
- Think of old TVs with the big, bulky screens. Those screens were CRTs.

Quick Tips:
- Remember that CRTs were the primary display technology before LCD and LED screens became common.

22 Easy

22 of 176Question permalink

Application +4 | -1

Which of the following ions has electronic configuration

\left[ {\mathrm{Ar}} \right]3{d^6}:

View answer and explanation

Correct option: D

{\mathrm{\;}_{27}}\mathrm{C}{\mathrm{o}^{3 + }}

Option D (

{\mathrm{\;}_{27}}\mathrm{C}{\mathrm{o}^{3 + }})

is correct because Cobalt (Co) has an atomic number of 27. Co³⁺ loses 3 electrons, resulting in the electronic configuration [Ar]3d⁶.

Solution approach

Correct Answer + Core Reason: ${\mathrm{\;}_{27}}\mathrm{C}{\mathrm{o}^{3 + }}$ . Cobalt (Co) loses 3 electrons to form Co³⁺, resulting in the electronic configuration [Ar]3d⁶.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each neutral atom.

Step 2: Determine the electronic configuration of each ion by removing the appropriate number of electrons.

Step 3: Compare the resulting electronic configurations with the given configuration [Ar]3d⁶.

Step 4: For Ni³⁺ (Z=27): Ni: [Ar] 3d⁸ 4s². Ni³⁺: [Ar] 3d⁷ (incorrect)\nFor Mn³⁺ (Z=25): Mn: [Ar] 3d⁵ 4s². Mn³⁺: [Ar] 3d⁴ (incorrect)\nFor Fe³⁺ (Z=26): Fe: [Ar] 3d⁶ 4s². Fe³⁺: [Ar] 3d⁵ (incorrect)\nFor Co³⁺ (Z=27): Co: [Ar] 3d⁷ 4s². Co³⁺: [Ar] 3d⁶ (correct)

Why other options are wrong:
- Option A: ${\mathrm{\;}_{27}}\mathrm{N}{\mathrm{i}^{3 + }}$ is incorrect because Ni³⁺ has the electronic configuration [Ar]3d⁷.
- Option B: ${\mathrm{\;}_{25}}\mathrm{M}{\mathrm{n}^{3 + }}$ is incorrect because Mn³⁺ has the electronic configuration [Ar]3d⁴.
- Option C: ${\mathrm{\;}_{26}}\mathrm{F}{\mathrm{e}^{3 + }}$ is incorrect because Fe³⁺ has the electronic configuration [Ar]3d⁵.

Key Concepts:
- Electronic configuration
- Ion formation

Memory Aids:
- Remember to remove electrons from the outermost shell (highest n) first.

Quick Tips:
- Always write the electronic configuration of the neutral atom before removing electrons to form the ion.

23 Easy

23 of 176Question permalink

Remembering +4 | -1

The minimum energy required to eject an electron from an atom is called :

View answer and explanation

Correct option: D

Work function

Option D (Work function) is correct because the work function is defined as the minimum energy required to eject an electron from an atom.

Solution approach

Correct Answer + Core Reason: Work function. The work function is the minimum energy needed to remove an electron from a solid.

Step-by-Step Solution:

Step 1: Define the problem: The question asks for the term that describes the minimum energy required to eject an electron from an atom.

Step 2: Recall the definition of work function: The work function ( $W$ ) is the minimum amount of energy required to remove an electron from a solid to a point immediately outside the solid surface.

Step 3: Relate the definition to the question: The definition directly answers the question.

Step 4: Conclusion: The correct answer is work function.

Why other options are wrong:
- Option A: Kinetic energy is the energy of motion, not the minimum energy required for ejection.
- Option B: Electrical energy is the energy associated with electric fields and charges, not specifically the minimum energy for electron ejection.
- Option C: Chemical energy is the energy stored in chemical bonds, not the minimum energy for electron ejection.
- Option D: Work function is the correct term as it is the minimum energy required to remove an electron from a solid.

Key Concepts:
- Work function
- Photoelectric effect

Memory Aids:
- Work function: Think of it as the "work" needed to "functionally" remove an electron.

Quick Tips:
- Remember the definition of work function in the context of the photoelectric effect.

24 Easy

24 of 176Question permalink

Understanding +4 | -1

\mathrm{M}{\mathrm{g}^{2 + }}

is isoelectrionic with

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{a}^ + }

Option C (

\mathrm{N}{\mathrm{a}^ + }

) is correct because both

\mathrm{M}{\mathrm{g}^{2 + }}

and

\mathrm{N}{\mathrm{a}^ + }

have 10 electrons, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}^ + }$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $\mathrm{M}{\mathrm{g}^{2 + }}$ . Magnesium (Mg) has 12 electrons. $\mathrm{M}{\mathrm{g}^{2 + }}$ has lost 2 electrons, so it has 12 - 2 = 10 electrons.

Step 2: Determine the number of electrons in each of the options:
- A. $\mathrm{C}{\mathrm{u}^{2 + }}$ : Copper (Cu) has 29 electrons. $\mathrm{C}{\mathrm{u}^{2 + }}$ has 29 - 2 = 27 electrons.
- B. $\mathrm{Z}{\mathrm{n}^{2 + }}$ : Zinc (Zn) has 30 electrons. $\mathrm{Z}{\mathrm{n}^{2 + }}$ has 30 - 2 = 28 electrons.
- C. $\mathrm{N}{\mathrm{a}^ + }$ : Sodium (Na) has 11 electrons. $\mathrm{N}{\mathrm{a}^ + }$ has 11 - 1 = 10 electrons.
- D. $\mathrm{C}{\mathrm{a}^{2 + }}$ : Calcium (Ca) has 20 electrons. $\mathrm{C}{\mathrm{a}^{2 + }}$ has 20 - 2 = 18 electrons.

Step 3: Identify the ion with the same number of electrons as $\mathrm{M}{\mathrm{g}^{2 + }}$ . $\mathrm{N}{\mathrm{a}^ + }$ has 10 electrons.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}^{2 + }}$ has 27 electrons, not 10.
- Option B: $\mathrm{Z}{\mathrm{n}^{2 + }}$ has 28 electrons, not 10.
- Option D: $\mathrm{C}{\mathrm{a}^{2 + }}$ has 18 electrons, not 10.

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.

25 Easy

25 of 176Question permalink

Application +4 | -1

According to (n+l) rule after completing 'np' level the electron enters to:

View answer and explanation

Correct option: B

\left( {n + 1} \right)s

Option B (

(n + 1)s

) is correct because according to the (n+l) rule, after filling the 'np' level, the next orbital to be filled is the (n+1)s orbital, as it has the lowest (n+l) value.

Solution approach

Correct Answer + Core Reason: $(n + 1)s$ . According to the (n+l) rule, the orbital with the lowest (n+l) value is filled first.

Step-by-Step Solution:

Step 1: Recall the (n+l) rule, which states that electrons first fill the orbital with the lowest (n+l) value. If two orbitals have the same (n+l) value, the electron fills the orbital with the lower n value first.

Step 2: Consider the given options after completing the 'np' level.

Step 3: After 'np', we need to compare the (n+l) values for (n-1)d, (n+1)s, nd, and (n+1)p.
(n-1)d: n-1 + 2 = n+1
(n+1)s: n+1 + 0 = n+1\nnd: n + 2 = n+2
(n+1)p: n+1 + 1 = n+2

Step 4: Since (n-1)d and (n+1)s both have n+l = n+1, we fill the one with the lower n value first. (n+1)s has a higher n value than (n-1)d, but the question is after 'np' level. So we compare (n+1)s with (n+1)p and nd. Since (n+1)s has the lowest (n+l) value, it is filled first.

Why other options are wrong:
- Option A: $(n - 1)d$ is incorrect because although it has the same (n+l) value as (n+1)s, the (n+1)s orbital is filled after the np orbital.
- Option C: nd is incorrect because it has a higher (n+l) value than (n+1)s.
- Option D: $(n + 1)p$ is incorrect because it has a higher (n+l) value than (n+1)s.

Key Concepts:
- (n+l) rule
- Aufbau principle

Memory Aids:
- Remember the (n+l) rule: lower (n+l) fills first.

Quick Tips:
- Always calculate the (n+l) values for each orbital to determine the filling order.

26 Easy

26 of 176Question permalink

Understanding +4 | -1

Isoelectronic pair among the following is

View answer and explanation

Correct option: B

Ar and

\mathrm{C}{\mathrm{a}^{2 + }}

Option B (Ar and

\mathrm{C}{\mathrm{a}^{2 + }}

) is correct because both species have the same number of electrons (18).

Solution approach

Correct Answer + Core Reason: Ar and $\mathrm{C}{\mathrm{a}^{2 + }}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Define isoelectronic species.
- Isoelectronic species have the same number of electrons.

Step 2: Determine the number of electrons in each species:
- Ca: 20 electrons
- K: 19 electrons
- Ar: 18 electrons
- $\mathrm{C}{\mathrm{a}^{2 + }}$ : 20 - 2 = 18 electrons

Step 3: Identify the isoelectronic pair.
- Ar and $\mathrm{C}{\mathrm{a}^{2 + }}$ both have 18 electrons.

Why other options are wrong:
- Option A: Ca and K have different numbers of electrons (20 and 19, respectively).
- Option C: K and $\mathrm{C}{\mathrm{a}^{2 + }}$ have different numbers of electrons (19 and 18, respectively).
- Option D: Ar and K have different numbers of electrons (18 and 19, respectively).

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

Memory Aids:
- Remember that positive ions have fewer electrons than their neutral atoms, and negative ions have more.

Quick Tips:
- To find the number of electrons in an ion, subtract the charge from the atomic number (for positive ions) or add the charge to the atomic number (for negative ions).

27 Easy

27 of 176Question permalink

Application +4 | -1

Magnetic quantum number for the last electron in sodium is:

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because the last electron in sodium (1s² 2s² 2p⁶ 3s¹) occupies the 3s orbital, which has an azimuthal quantum number (l) of 0. Therefore, the magnetic quantum number (m) is also 0.

Solution approach

Correct Answer + Core Reason: Zero. The last electron in sodium is in the 3s orbital, which has a magnetic quantum number of 0.

Step-by-Step Solution:

Step 1: Write the electronic configuration of sodium (Na): $1s^2 2s^2 2p^6 3s^1$ .

Step 2: Identify the last electron. It is in the 3s orbital.

Step 3: Determine the azimuthal quantum number ( $l$ ) for the s orbital. For s orbitals, $l = 0$ .

Step 4: Determine the magnetic quantum number ( $m$ ) for $l = 0$ . The magnetic quantum number can range from $-l$ to $+l$ , so for $l = 0$ , $m = 0$ .

Why other options are wrong:
- Option A: 3 is incorrect because the magnetic quantum number cannot be 3 for the last electron in sodium.
- Option B: 1 is incorrect because the magnetic quantum number cannot be 1 for the last electron in sodium.
- Option C: 2 is incorrect because the magnetic quantum number cannot be 2 for the last electron in sodium.

Key Concepts:
- Electronic configuration
- Quantum numbers (principal, azimuthal, magnetic)
- s orbitals

Memory Aids:
- For s orbitals, $l = 0$ and $m = 0$ .

Quick Tips:
- Always write the electronic configuration first to identify the orbital of the last electron.

28 Easy

28 of 176Question permalink

Remembering +4 | -1

Different lines in Lyman series of hydrogen spectrum lie in …. Region

View answer and explanation

Correct option: A

Ultraviolet

Option A (Ultraviolet) is correct because the Lyman series corresponds to transitions to the n=1 energy level, which results in the emission of ultraviolet photons.

Solution approach

Correct Answer + Core Reason: Ultraviolet. Lyman series lies in the ultraviolet region.

Step-by-Step Solution:

Step 1: Recall the regions of the electromagnetic spectrum for different series in the hydrogen spectrum:
- Lyman series: Ultraviolet region
- Balmer series: Visible region
- Paschen, Brackett, and Pfund series: Infrared region

Step 2: Identify the region for the Lyman series. The Lyman series lies in the ultraviolet region.

Why other options are wrong:
- Option B: Infrared region is for Paschen, Brackett, and Pfund series.
- Option C: Visible region is for the Balmer series.
- Option D: Far infrared is for transitions to very high n values.

Key Concepts:
- Hydrogen spectrum
- Lyman series
- Electromagnetic spectrum

Memory Aids:
- Remember the regions for each series in the hydrogen spectrum.

Quick Tips:
- Memorize the regions of the electromagnetic spectrum for each series in the hydrogen spectrum.

29 Easy

29 of 176Question permalink

Remembering +4 | -1

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

View answer and explanation

Correct option: B

Emission spectra

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

Solution approach

Correct Answer + Core Reason: Emission spectra. When an electron transitions from a higher to a lower energy level, it emits energy in the form of photons, resulting in emission spectra.

Step-by-Step Solution:

Step 1: Understand the concept of electron transitions in atoms.

Step 2: When an electron moves from a higher energy orbit to a lower energy orbit, it releases energy.

Step 3: This released energy is in the form of a photon.

Step 4: The collection of these emitted photons at specific wavelengths forms the emission spectra.

Why other options are wrong:
- Option A: Absorption spectra: Absorption spectra are produced when an electron absorbs energy and moves from a lower to a higher energy level.
- Option C: $\alpha -$ particle: Alpha particles are helium nuclei and are not produced during electron transitions.
- Option D: None of these: Emission spectra are indeed produced.

Key Concepts:
- Electron transitions
- Emission spectra
- Absorption spectra

Memory Aids:
- Emission: Electron moves from higher to lower energy level (Emits energy).
- Absorption: Electron moves from lower to higher energy level (Absorbs energy).

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

30 Easy

30 of 176Question permalink

Application +4 | -1

The frequency of a green light is

6 \times {10^{14}}\mathrm{Hz}.

Its wavelength is:

View answer and explanation

Correct option: A

500 nm

Option A (500 nm) is correct because wavelength is the speed of light divided by the frequency, and using the given frequency yields a wavelength of 500 nm.

Solution approach

Correct Answer + Core Reason: 500 nm. The wavelength is calculated using the formula $\lambda = \frac{c}{f}$ .

Step-by-Step Solution:

Step 1: Recall the relationship between frequency, wavelength, and the speed of light: $c = \lambda f$ , where $c$ is the speed of light ( $3 \times 10^8 \text{ m/s}$ ), $\lambda$ is the wavelength, and $f$ is the frequency.

Step 2: Rearrange the formula to solve for wavelength: $\lambda = \frac{c}{f}$ .

Step 3: Substitute the given values: $\lambda = \frac{3 \times 10^8 \text{ m/s}}{6 \times 10^{14} \text{ Hz}}$ .

Step 4: Calculate the wavelength: $\lambda = 0.5 \times 10^{-6} \text{ m} = 500 \times 10^{-9} \text{ m} = 500 \text{ nm}$ .

Why other options are wrong:
- Option B: 5 nm - Incorrect, this is significantly smaller than the correct wavelength.
- Option C: 50,000 nm - Incorrect, this is significantly larger than the correct wavelength.
- Option D: None of these - Incorrect, the correct wavelength is 500 nm.

Key Concepts:
- Relationship between frequency, wavelength, and speed of light
- Electromagnetic spectrum

Memory Aids:
- Remember the formula $c = \lambda f$ .

Quick Tips:
- Ensure the units are consistent before performing calculations.

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