STRUCTURE OF ATOM MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

Remembering +4 | -1

The region in the electromagnetic spectrum where the Balmer series lines appear is

View answer and explanation

Correct option: D

visible

Option D (visible) is correct because the Balmer series lines are produced by electron transitions that emit photons in the visible region of the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: visible. The Balmer series corresponds to electron transitions ending in the n=2 energy level, which emits photons in the visible region.

Step-by-Step Solution:

Step 1: Recall the Balmer series definition. The Balmer series consists of spectral lines resulting from the transitions of an electron from higher energy levels to the n=2 energy level in a hydrogen atom.

Step 2: Relate the Balmer series to the electromagnetic spectrum. The wavelengths of the emitted photons in the Balmer series fall within the visible region of the electromagnetic spectrum.

Step 3: Conclusion: Therefore, the region in the electromagnetic spectrum where the Balmer series lines appear is the visible region.

Why other options are wrong:
- Option A (Infrared): Infrared radiation has longer wavelengths and lower energy than visible light. Transitions to higher energy levels (lower n values) would be required for infrared emission in the hydrogen atom.
- Option B (ultraviolet): Ultraviolet radiation has shorter wavelengths and higher energy than visible light. The Lyman series (transitions to n=1) falls in the ultraviolet region.
- Option C (microwave): Microwaves have much longer wavelengths and lower energy than visible light. These are associated with rotational transitions in molecules, not electronic transitions in atoms.

Key Concepts:
- Balmer Series
- Electromagnetic Spectrum
- Atomic Spectra

Memory Aids:
- "Balmer is Visible": A mnemonic to remember that the Balmer series is in the visible region.

Quick Tips:
- Remember the different series (Lyman, Balmer, Paschen, Brackett, Pfund) and their corresponding regions in the electromagnetic spectrum.

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Analyzing +4 | -1

If Rutherford would have used

\beta

-particles instead of

\alpha

-particles in his Gold-leaf experiment, which of the following observations would definitely not have been made:

View answer and explanation

Correct option: A

\beta

-particles scattering beyond the gold foil.

Option A is correct because beta particles are much lighter than alpha particles and would not be significantly scattered beyond the gold foil due to the relatively small mass of electrons in the atoms.

Solution approach

Correct Answer + Core Reason: $\beta$ -particles scattering beyond the gold foil. Beta particles are much lighter than alpha particles. Therefore, they would not be significantly scattered beyond the gold foil due to the relatively small mass of electrons in the atoms.

Step-by-Step Solution:

Step 1: Recall Rutherford's gold foil experiment. Alpha particles were used to bombard a thin gold foil.

Step 2: Consider the mass and charge of alpha and beta particles. Alpha particles are heavy and positively charged. Beta particles are light (electrons) and negatively charged.

Step 3: Analyze the scattering pattern with alpha particles. Most alpha particles passed through undeflected, some were deflected at small angles, and very few were deflected at large angles.

Step 4: Predict the scattering pattern with beta particles. Since beta particles are much lighter, they would mostly pass through undeflected. The scattering would be primarily due to interactions with the electrons in the gold atoms. Significant scattering beyond the foil would not be observed.

Why other options are wrong:
- Option B: Majority of $\beta$ -particles passing undeflected through the foil would still be observed.
- Option C: Very few $\beta$ -particles getting absorbed would still be observed.
- Option D: Wave nature of electron is a separate concept and not directly related to the scattering experiment in this context.

Key Concepts:
- Rutherford's gold foil experiment
- Alpha particles
- Beta particles
- Scattering

Memory Aids:
- Alpha particles are heavy, beta particles are light.

Quick Tips:
- Consider the mass and charge of the particles when analyzing scattering experiments.

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Application +4 | -1

The ratio of radius of III and IV Bohr's orbits in hydrogen atom is:

View answer and explanation

Correct option: C

9∶16

Option C (9∶16) is correct because the radius of the nth Bohr orbit is proportional to

n^2

, so the ratio of the radii is the ratio of the squares of the orbit numbers.

Solution approach

Correct Answer + Core Reason: 9∶16. The radius of the nth Bohr orbit is proportional to $n^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n \propto n^2$ .

Step 2: We want to find the ratio of the radius of the third Bohr orbit (n=3) to the radius of the fourth Bohr orbit (n=4).

Step 3: Calculate the ratio: $\frac{r_3}{r_4} = \frac{3^2}{4^2} = \frac{9}{16}$ .

Step 4: Therefore, the ratio of the radii is 9:16.

Why other options are wrong:
- Option A: This is incorrect because it takes the direct ratio of the orbit numbers instead of the ratio of their squares.
- Option B: This is incorrect and doesn't follow the correct formula for the radius of the nth Bohr orbit.
- Option D: This is incorrect and doesn't follow the correct formula for the radius of the nth Bohr orbit.

Key Concepts:
- Bohr's model of the atom
- Radius of the nth Bohr orbit: $r_n \propto n^2$

Memory Aids:
- Remember that the radius increases with the square of the orbit number.

Quick Tips:
- Always remember the formula for the radius of the nth Bohr orbit. This type of question is a direct application of the formula.

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Application +4 | -1

Which one of the following set of quantum numbers is possible for electron in the ground state of an atom with atomic number 19?

View answer and explanation

Correct option: D

n = 4,\;l = 0,\;m = 0

Option D (

n = 4,\;l = 0,\;m = 0

) is correct because the ground state electronic configuration of an atom with atomic number 19 (Potassium) is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹, and the last electron occupies the 4s orbital, which has n=4, l=0, and m=0.

Solution approach

Correct Answer + Core Reason: $n = 4,\;l = 0,\;m = 0$ . The last electron of potassium (Z=19) occupies the 4s orbital.

Step-by-Step Solution:

Step 1: Determine the element with atomic number 19. It is potassium (K).

Step 2: Write the ground state electronic configuration of potassium: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.

Step 3: Identify the quantum numbers for the last electron (4s¹).\nn = 4 (principal quantum number)\nl = 0 (azimuthal quantum number for s orbital)\nm = 0 (magnetic quantum number for l=0)

Why other options are wrong:
- Option A: $n = 2,\;l = 0,\;m = 0$ is incorrect because this corresponds to a 2s electron, which is not the last electron in potassium.
- Option B: $n = 2,\;l = 1,\;m = 0$ is incorrect because this corresponds to a 2p electron, which is not the last electron in potassium.
- Option C: $n = 3,\;l = 1,\;m = - 1$ is incorrect because this corresponds to a 3p electron, which is not the last electron in potassium.

Key Concepts:
- Electronic configuration
- Quantum numbers

Memory Aids:
- Remember the quantum numbers for s, p, d, and f orbitals: l=0, 1, 2, and 3, respectively.

Quick Tips:
- Always write the complete electronic configuration to determine the quantum numbers of the last electron.

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Remembering +4 | -1

The momentum of a photon is

p,

the energy associated with photon is given by:

View answer and explanation

Correct option: B

\frac{E}{p}

Option B (

\frac{E}{p}

) is correct because the energy of a photon is related to its momentum by the equation

E = pc

, so

c = \frac{E}{p}

, and since the question is asking for the energy associated with the photon, the correct answer is

\frac{E}{p}

.

Solution approach

Correct Answer + Core Reason: $\frac{E}{p}$ . The energy associated with the photon is given by the ratio of energy to momentum.

Step-by-Step Solution:

Step 1: The question asks for the relationship between the energy and momentum of a photon.

Step 2: Recall the relationship between energy, momentum, and the speed of light for a photon: $E = pc$ , where:
- $E$ is the energy of the photon.
- $p$ is the momentum of the photon.
- $c$ is the speed of light.

Step 3: Solve for $c$ (the speed associated with the photon): $c = \frac{E}{p}$ .

Step 4: Therefore, the energy associated with the photon is given by $\frac{E}{p}$ .

Why other options are wrong:
- **Option A ( $\frac{p}{E}$ ): This is incorrect. This would give the inverse of the speed of light.
- Option C ( $p \cdot E$ ): This is incorrect. The energy and momentum are related by division, not multiplication.
- Option D ( $\sqrt{\frac{E}{p}}$ ): This is incorrect. The relationship is linear, not a square root.

Key Concepts:
- Photon energy
- Photon momentum
- Speed of light

Memory Aids:**
- Remember the relationship $E = pc$ .

Quick Tips:
- The speed of light is the constant of proportionality between energy and momentum for a photon.

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Understanding +4 | -1

The formation of energy bonds in solids are in accordance with

View answer and explanation

Correct option: B

Bohr’s theory

Option B (Bohr’s theory) is correct because Bohr's theory explains the discrete energy levels and transitions of electrons, which leads to the formation of energy bands in solids.

Solution approach

Correct Answer + Core Reason: Bohr’s theory. Bohr's theory explains the discrete energy levels and transitions of electrons, which leads to the formation of energy bands in solids.

Step-by-Step Solution:

Step 1: Consider the formation of energy bands in solids.

Step 2: Recall that Bohr's theory describes the quantized energy levels of electrons in atoms.

Step 3: In solids, the interaction between atoms causes these discrete energy levels to broaden into energy bands.

Step 4: Therefore, the formation of energy bands is in accordance with Bohr's theory.

Why other options are wrong:
- Option A: Heisenberg's uncertainty principle: This principle deals with the uncertainty in determining position and momentum simultaneously, not directly related to energy band formation.
- Option C: Ohm’s law: This law relates voltage, current, and resistance in a circuit, not related to energy band formation.
- Option D: Rutherford’s atomic model: This model describes the structure of the atom but does not explain energy levels or band formation.

Key Concepts:
- Bohr's theory of atomic structure
- Energy band formation in solids

Memory Aids:
- Bohr's theory explains quantized energy levels, which broaden into bands in solids.

Quick Tips:
- Remember that energy band formation is a consequence of the interaction between atoms in a solid, which is related to Bohr's atomic model.

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Understanding +4 | -1

Electron density in the YZ plane of

3{d_{{x^2} - {y^2}}}

orbital is

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because the

3d_{x^2-y^2}

orbital lies along the x and y axes; therefore, the electron density in the YZ plane, which is perpendicular to the x-axis, is zero.

Solution approach

Correct Answer + Core Reason: Zero. The $3d_{x^2-y^2}$ orbital has lobes along the x and y axes, so the electron density in the YZ plane is zero.

Step-by-Step Solution:

Step 1: Visualize the shape of the $3d_{x^2-y^2}$ orbital.

Step 2: The $3d_{x^2-y^2}$ orbital has four lobes, with two lobes lying along the x-axis and two lobes lying along the y-axis.

Step 3: The YZ plane is the plane where x = 0. Since the lobes of the $3d_{x^2-y^2}$ orbital lie along the x and y axes, the electron density in the YZ plane is zero.

Why other options are wrong:
- Option B: 0.5 is incorrect because the electron density is not uniformly distributed and is zero in the YZ plane.
- Option C: 0.75 is incorrect because the electron density is not uniformly distributed and is zero in the YZ plane.
- Option D: 0.9 is incorrect because the electron density is not uniformly distributed and is zero in the YZ plane.

Key Concepts:
- Shapes of d-orbitals
- Electron density
- Coordinate planes

Memory Aids:
- Remember the shapes of the d-orbitals and their orientation in space.

Quick Tips:
- Visualize the orbitals in 3D space to determine electron density in different planes.

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Understanding +4 | -1

A certain orbital has no angular nodes and two radial nodes. The orbital is

View answer and explanation

Correct option: B

3s

Option B (

3s

) is correct because a 3s orbital has two radial nodes and zero angular nodes.

Solution approach

Correct Answer + Core Reason: $3s$ . A 3s orbital has 2 radial nodes and 0 angular nodes.

Step-by-Step Solution:

Step 1: Define radial and angular nodes: Radial nodes are spherical surfaces where the probability of finding an electron is zero. Angular nodes are planar or conical surfaces where the probability of finding an electron is zero.

Step 2: Recall the formulas for calculating the number of nodes:
- Number of radial nodes = $n - l - 1$ , where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.
- Number of angular nodes = $l$

Step 3: Analyze the given information: We are given that the orbital has no angular nodes ( $l = 0$ ) and two radial nodes.

Step 4: Determine the orbital: Since $l = 0$ , the orbital is an s orbital. The number of radial nodes is 2, so $n - l - 1 = 2$ . Substituting $l = 0$ , we get $n - 0 - 1 = 2$ , which means $n = 3$ .

Step 5: Final Answer: $3s$

Why other options are wrong:
- Option A: $2s$ : A 2s orbital has 0 angular nodes and 1 radial node ( $2 - 0 - 1 = 1$ ).
- Option C: $3p$ : A 3p orbital has 1 angular node and 1 radial node ( $3 - 1 - 1 = 1$ ).
- Option D: $2p$ : A 2p orbital has 1 angular node and 0 radial nodes ( $2 - 1 - 1 = 0$ ).

Key Concepts:
- Radial nodes
- Angular nodes
- Quantum numbers (n and l)

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

Quick Tips:
- Remember the formulas for calculating the number of radial and angular nodes. Know the values of l for s, p, d, and f orbitals (0, 1, 2, 3 respectively).

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Remembering +4 | -1

Which of the following conditions is incorrect for a well behaved wave function

(\psi )

View answer and explanation

Correct option: C

\psi

must be infinite

Option C (

\psi

must be infinite) is correct because a well-behaved wave function must be finite, not infinite.

Solution approach

Correct Answer + Core Reason: $\psi$ must be infinite. A well-behaved wave function must be finite, single-valued, and continuous.

Step-by-Step Solution:

Step 1: Recall the properties of a well-behaved wave function: A well-behaved wave function must be:
- Finite: The wave function must have a finite value at all points in space.
- Single-valued: The wave function must have only one value at any given point in space.
- Continuous: The wave function must not have any breaks or jumps.

Step 2: Compare the given conditions with the properties of a well-behaved wave function.

Why other options are wrong:
- Option A: $\psi$ must be finite is a correct condition for a well-behaved wave function.
- Option B: $\psi$ must be single valued is a correct condition for a well-behaved wave function.
- Option C: $\psi$ must be infinite is an incorrect condition. A well-behaved wave function must be finite, not infinite, because an infinite value would imply an infinite probability of finding the particle at that point, which is not physically possible.
- Option D: $\psi$ must be continuous is a correct condition for a well-behaved wave function.

Key Concepts:
- Wave function
- Properties of a well-behaved wave function
- Quantum mechanics

Memory Aids:
- Think of a well-behaved function as being "nice": finite, single, and continuous.

Quick Tips:
- Remember the three main properties of a well-behaved wave function: finite, single-valued, and continuous.

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Application +4 | -1

If the radius of first Bohr's orbit be

{a_0}

, then the radius of third Bohr's orbit would be:

View answer and explanation

Correct option: C

9 \times {a_0}

Option C (

9 \times {a_0}

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

n^2

.

Solution approach

Correct Answer + Core Reason: $9 \times {a_0}$ . The radius of the nth Bohr orbit is proportional to $n^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n = a_0 n^2$ , where $a_0$ is the radius of the first Bohr orbit and n is the orbit number.

Step 2: We are given that the radius of the first Bohr orbit is $a_0$ . We want to find the radius of the third Bohr orbit, so $n = 3$ .

Step 3: Substitute $n = 3$ into the formula: $r_3 = a_0 (3)^2 = 9 a_0$ .

Step 4: Therefore, the radius of the third Bohr orbit is $9a_0$ .

Why other options are wrong:
- Option A: This is incorrect because it assumes a linear relationship between the orbit number and the radius, rather than a quadratic relationship.
- Option B: This is incorrect and doesn't follow the correct formula for the radius of the nth Bohr orbit.
- Option D: This is incorrect because it calculates the inverse square of the orbit number, rather than the square.

Key Concepts:
- Bohr's model of the atom
- Radius of the nth Bohr orbit: $r_n = a_0 n^2$

Memory Aids:
- Remember that the radius increases with the square of the orbit number.

Quick Tips:
- Always remember the formula for the radius of the nth Bohr orbit. This type of question is a direct application of the formula.

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Understanding +4 | -1

Among the following, the correct statement about cathode ray discharge tube is

View answer and explanation

Correct option: B

in the absence of external electrical or magnetic field, cathode rays travel in straight lines.

Option B is correct because in the absence of external fields, cathode rays, being streams of electrons, travel in straight lines.

Solution approach

Correct Answer + Core Reason: in the absence of external electrical or magnetic field, cathode rays travel in straight lines. Cathode rays are streams of electrons and travel in straight lines unless deflected by external fields.

Step-by-Step Solution:

Step 1: Cathode rays are streams of electrons emitted from the cathode in a discharge tube.

Step 2: In the absence of external electric or magnetic fields, these electrons travel in straight lines due to their inertia.

Step 3: Applying an electric or magnetic field causes the cathode rays to deflect, demonstrating their charged nature.

Why other options are wrong:
- Option A: The electrical discharge is observed at low pressure and high voltage, not high pressure and low voltage.
- Option C: The characteristics of cathode rays do not depend on the material of the electrodes. They depend on the gas used in the tube.
- Option D: The characteristics of cathode rays do not depend on the gas present in the cathode ray tube. They are fundamental properties of electrons.

Key Concepts:
- Cathode rays
- Properties of electrons
- Discharge tube experiment

Memory Aids:
- Cathode rays travel in straight lines unless acted upon by a force.

Quick Tips:
- Remember the conditions (low pressure, high voltage) for observing cathode rays.

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Remembering +4 | -1

If ‘r’ is the radius of first orbit, the radius of nth orbit of H-atom is given by

View answer and explanation

Correct option: A

r{n^2}

Option A (

r{n^2}

) is correct because the radius of the nth orbit in a hydrogen atom is proportional to

n^2

, where n is the principal quantum number.

Solution approach

Correct Answer + Core Reason: $r{n^2}$ . The radius of the nth orbit of a hydrogen atom is proportional to $n^2$ .

Step-by-Step Solution:

Step 1: Bohr's Model: According to Bohr's model of the hydrogen atom, the radius of the nth orbit is given by the formula:
$r_n = r_1 n^2$ , where $r_1$ is the radius of the first orbit (Bohr radius) and $n$ is the principal quantum number (n = 1, 2, 3, ...).

Step 2: Radius of nth Orbit: If 'r' is the radius of the first orbit ( $r = r_1$ ), then the radius of the nth orbit is $r_n = r n^2$ .

Why other options are wrong:
- Option B: $rn$ . This is incorrect. The radius is proportional to $n^2$ , not $n$ .
- Option C: $\frac{r}{n}$ . This is incorrect. The radius increases with $n$ , not decreases.
- Option D: ${r^2}{n^2}$ . This is incorrect. The radius is proportional to $n^2$ , not $r^2 n^2$ .

Key Concepts:
- Bohr's model of the hydrogen atom
- Radius of nth orbit
- Principal quantum number

Memory Aids:
- Radius increases with the square of the orbit number ( $n^2$ ).

Quick Tips:
- Remember the formula for the radius of the nth orbit: $r_n = r_1 n^2$ .

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Application +4 | -1

The Bohr’s orbit radius for the hydrogen atom (n=1) is approximately 0.53 Å. The radius for the first excited state (n=2) orbit is

View answer and explanation

Correct option: C

2.12 Å

Option C (2.12 Å) is correct because the radius of the nth orbit is proportional to

n^2

, so the radius of the first excited state (n=2) is 4 times the radius of the ground state (n=1).

Solution approach

Correct Answer + Core Reason: 2.12 Å. The radius of the nth orbit is proportional to $n^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n = a_0 n^2$ , where $a_0$ is a constant and n is the orbit number.

Step 2: We are given that the Bohr radius for the hydrogen atom (n=1) is approximately 0.53 Å.

Step 3: We want to find the radius for the first excited state (n=2).

Step 4: Substitute $n = 2$ into the formula: $r_2 = 0.53 \times (2)^2 = 0.53 \times 4 = 2.12$ Å.

Why other options are wrong:
- Option A: This is incorrect because it divides the ground state radius by 2 instead of multiplying by 4.
- Option B: This is incorrect and doesn't follow the correct formula for the radius of the nth Bohr orbit.
- Option D: This is incorrect and doesn't follow the correct formula for the radius of the nth Bohr orbit.

Key Concepts:
- Bohr's model of the atom
- Radius of the nth Bohr orbit: $r_n \propto n^2$

Memory Aids:
- Remember that the radius increases with the square of the orbit number.

Quick Tips:
- Always remember the formula for the radius of the nth Bohr orbit. This type of question is a direct application of the formula.

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Analyzing +4 | -1

Assertion: In absorption spectrum,

n_f > n_i

, while in emission spectrum,

n_f > n_i

Reason: Energy is limited in absorption spectrum, that is, electron moves from lower to higher orbit and energy is absorbed in emission spectrum, because election jumps from higher to lower orbit.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

Option D (If both the Assertion and Reason are incorrect) is correct because the assertion incorrectly states the relationship between initial and final energy levels in absorption and emission spectra, and the reason provides an inaccurate explanation of these processes.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. The assertion and reason both contain incorrect information about absorption and emission spectra.

Step-by-Step Solution:

Step 1: Analyze the Assertion: In absorption spectrum, an electron transitions from a lower energy level to a higher energy level, so $n_f > n_i$ is correct. However, in emission spectrum, an electron transitions from a higher energy level to a lower energy level, so $n_f < n_i$ . Therefore, the assertion is incorrect.

Step 2: Analyze the Reason: Energy is absorbed in absorption spectrum because the electron moves from a lower to a higher orbit. Energy is emitted in emission spectrum because the electron jumps from a higher to a lower orbit. The reason correctly describes the energy changes in absorption and emission. However, since the assertion is incorrect, the reason cannot be the correct explanation.

Step 3: Since the assertion is incorrect and the reason, while partially correct in its description of energy changes, is not a valid explanation for the incorrect assertion, the correct answer is that both the assertion and reason are incorrect.

Why other options are wrong:
- Option A: This is incorrect because the assertion is false.
- Option B: This is incorrect because the assertion is false.
- Option C: This is incorrect because the reason is also incorrect in the context of the false assertion.

Key Concepts:
- Absorption spectrum
- Emission spectrum
- Energy levels and transitions

Memory Aids:
- Remember that absorption involves moving to a higher energy level, and emission involves moving to a lower energy level.

Quick Tips:
- Carefully analyze the definitions of absorption and emission spectra.
- Pay attention to the direction of electron transitions.

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Remembering +4 | -1

The radiation having maximum wavelength is

View answer and explanation

Correct option: B

Radio waves

Option B (Radio waves) is correct because radio waves have the longest wavelength in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: Radio waves. Radio waves have the longest wavelength.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum.

Step 2: The electromagnetic spectrum, in order of increasing frequency (decreasing wavelength) is:
- Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.

Step 3: Identify the radiation with the maximum wavelength.

Step 4: Radio waves have the maximum wavelength.

Why other options are wrong:
- Option A: UV rays have shorter wavelengths than radio waves.
- Option C: X-rays have shorter wavelengths than radio waves.
- Option D: IR rays have shorter wavelengths than radio waves.

Key Concepts:
- Electromagnetic spectrum
- Wavelength and frequency

Memory Aids:
- Remember the order of the electromagnetic spectrum.

Quick Tips:
- Know the order of the electromagnetic spectrum by wavelength and frequency.

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Application +4 | -1

What is the minimum energy that photons must posses in order to produce photoelectric effect with platinum metal? The threshold frequency for platinum is

1.3 \times {10^{15}}{s^{ - 1}}

View answer and explanation

Correct option: D

8.6 \times {10^{ - 12}}\,erg

Option D (

8.6 \times {10^{ - 12}}\,erg

) is correct because it is calculated using the formula E = h*v, where h is Planck's constant and v is the threshold frequency.

Solution approach

Correct Answer + Core Reason: $8.6 \times {10^{ - 12}}\,erg$ . The minimum energy is calculated using Planck's equation.

Step-by-Step Solution:

Step 1: Identify the given information: Threshold frequency ( $v$ ) = $1.3 \times {10^{15}}{s^{ - 1}}$ . We need to find the minimum energy (E).

Step 2: Recall Planck's equation: $E = h \times v$ , where $h$ is Planck's constant.

Step 3: The value of Planck's constant (h) in cgs units is $6.626 \times {10^{ - 27}} erg \cdot s$ .

Step 4: Substitute the values into the equation: $E = (6.626 \times {10^{ - 27}} erg \cdot s) \times (1.3 \times {10^{15}}{s^{ - 1}})$

Step 5: Calculate the energy: $E = 8.6138 \times {10^{ - 12}} erg$

Step 6: Round to two significant figures: $E \approx 8.6 \times {10^{ - 12}} erg$

Final Answer: $8.6 \times {10^{ - 12}} erg$

Why other options are wrong:
- Option A: This value is incorrect because it is significantly smaller than the correct energy value. It might be a result of incorrect calculation or using the wrong value for Planck's constant.
- Option B: This value is incorrect. It's likely a result of a calculation error, possibly involving incorrect powers of 10.
- Option C: This value is incorrect. It's likely a result of a calculation error, possibly involving incorrect powers of 10.
- Option D: This is the correct answer.

Key Concepts:
- Photoelectric effect
- Planck's equation: $E = h \times v$
- Threshold frequency

Memory Aids:
- Remember Planck's equation: E = hv

Quick Tips:
- Always use the correct units for Planck's constant (J.s or erg.s).
- Pay attention to the powers of 10 when performing calculations.

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Analyzing +4 | -1

The ratio of slopes of

K_{\max } v s . v

and

V_0 v s . v

curves in the photoelectric effect gives

(v=

frequency,

K_{\max }=

maximum kinetic energy,

V_0=

stopping potential) :

View answer and explanation

Correct option: A

charge of electron

Option A (charge of electron) is correct. The ratio of the slopes gives the charge of the electron as it relates the maximum kinetic energy and stopping potential to the frequency.

Solution approach

Correct Answer + Core Reason: charge of electron. The ratio of the slopes of the Kmax vs. v and V0 vs. v curves gives the charge of the electron.

Step-by-Step Solution:

Step 1: Write the photoelectric equation: $K_{max} = h\nu - \phi$ , where $K_{max}$ is the maximum kinetic energy, $h$ is Planck's constant, $\nu$ is the frequency, and $\phi$ is the work function.

Step 2: The slope of the $K_{max}$ vs. $\nu$ curve is $h$ .

Step 3: Write the relation between stopping potential and maximum kinetic energy: $K_{max} = eV_0$ , where $e$ is the charge of the electron and $V_0$ is the stopping potential.

Step 4: Therefore, $eV_0 = h\nu - \phi$ , so $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$ .

Step 5: The slope of the $V_0$ vs. $\nu$ curve is $\frac{h}{e}$ .

Step 6: The ratio of the slopes is $\frac{h}{\frac{h}{e}} = e$ .

Step 7: Therefore, the ratio of the slopes is the charge of the electron.

Why other options are wrong:
- Option B: Planck's constant is the slope of the Kmax vs frequency curve, but not the ratio of the slopes.
- Option C: Work function is related to the intercept of the V0 vs frequency curve, not the ratio of slopes.
- Option D: The ratio of Planck's constant to electronic charge is the slope of the V0 vs frequency curve, not the ratio of the slopes of the two curves.

Key Concepts:
- Photoelectric effect
- Maximum kinetic energy
- Stopping potential
- Planck's constant
- Work function

Memory Aids:
- $K_{max} = h\nu - \phi$
- $K_{max} = eV_0$

Quick Tips:
- Remember the photoelectric equation and the relationship between stopping potential and maximum kinetic energy.

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Understanding +4 | -1

During photoelectric effect, the kinetic energy of the electron ejected from metal surface depends on

View answer and explanation

Correct option: C

The frequency of incident light

Option C is correct. The kinetic energy of the ejected electron in the photoelectric effect depends on the frequency of the incident light.

Solution approach

Correct Answer + Core Reason: The frequency of incident light. The kinetic energy of ejected electrons depends on the frequency of the incident light according to the photoelectric equation.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $KE = h\nu - \phi$ , where $KE$ is the kinetic energy, $h\nu$ is the energy of the incident photon, and $\phi$ is the work function.

Step 2: Analyze the equation: The kinetic energy depends on the frequency ( $v$ ) of the incident light.

Step 3: Consider intensity and brightness: Intensity and brightness relate to the number of photons, not the kinetic energy of individual electrons.

Step 4: Conclusion: The kinetic energy of the electron ejected from the metal surface depends on the frequency of incident light.

Why other options are wrong:
- Option A: The intensity of incident light: Incorrect because intensity affects the number of electrons ejected, not their kinetic energy.
- Option B: The brightness of the incident light: Incorrect because brightness is related to intensity and affects the number of electrons ejected, not their kinetic energy.
- Option D: Both 1 and 2: Incorrect because neither intensity nor brightness directly affects the kinetic energy of the ejected electrons.

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

Memory Aids:
- Frequency determines kinetic energy, intensity determines the number of electrons.

Quick Tips:
- Remember the photoelectric equation and the roles of frequency and intensity.

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Understanding +4 | -1

For the Balmer series in the spectrum of

H

atom,

\overline v = {R_H}\left\{ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right\}

the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer

{n_1}

is equal to
(III) The lines of longest wavelength corresponds to

{n_2} = 3

(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines

View answer and explanation

Correct option: D

I, II, III

Option D (I, II, III) is correct because statements I, II, and III accurately describe the Balmer series.

Solution approach

Correct Answer + Core Reason: I, II, III. Statements I, II, and III are correct descriptions of the Balmer series.

Step-by-Step Solution:

Step 1: Understand the Balmer series: The Balmer series is a set of spectral lines in the visible region of the hydrogen spectrum, corresponding to transitions where the electron falls to the $n_1 = 2$ energy level.

Step 2: Analyze statement I: As wavelength decreases, the lines in the series converge. This is true because as $n_2$ increases, the energy difference between levels decreases, leading to smaller wavelength differences and convergence.

Step 3: Analyze statement II: The integer $n_1$ is equal to 2. This is true because the Balmer series corresponds to transitions to the $n_1 = 2$ energy level.

Step 4: Analyze statement III: The lines of longest wavelength corresponds to $n_2 = 3$ . This is true because the smallest energy transition (and thus longest wavelength) occurs when the electron transitions from $n_2 = 3$ to $n_1 = 2$ .

Step 5: Analyze statement IV: The ionization energy of hydrogen can be calculated from wave number of these lines. This is false. The ionization energy is the energy required to remove an electron from the ground state ( $n=1$ ) to infinity. The Balmer series involves transitions to $n=2$ , not the ground state. The Lyman series is used to calculate ionization energy.

Step 6: Final Answer: I, II, III

Why other options are wrong:
- Option A: I, II, IV: Statement IV is incorrect.
- Option B: II, III, IV: Statement IV is incorrect.
- Option C: I, III, IV: Statement IV is incorrect.

Key Concepts:
- Balmer series
- Hydrogen spectrum
- Spectral lines
- Ionization energy

Memory Aids:
- Balmer: Visible, $n_1 = 2$

Quick Tips:
- Remember the different series in the hydrogen spectrum (Lyman, Balmer, Paschen, etc.) and their corresponding $n_1$ values.

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Application +4 | -1

The values of four quantum numbers of valence electron of an element are

n = 4,\;l = 0,\;m = 0\;\mathrm{and}\;s = + \frac{1}{2}\;

The element is

View answer and explanation

Correct option: A

K

Option A (K) is correct because the electronic configuration described corresponds to the valence electron of Potassium (K), which is 4s¹.

Solution approach

Correct Answer + Core Reason: K. The quantum numbers indicate the valence electron is in the 4s orbital.

Step-by-Step Solution:

Step 1: Analyze the given quantum numbers: $n = 4$ , $l = 0$ , $m = 0$ , and $s = +\frac{1}{2}$ .

Step 2: Determine the orbital based on the quantum numbers:
- $n = 4$ indicates the electron is in the 4th energy level.
- $l = 0$ indicates the electron is in an $s$ orbital.
- $m = 0$ is consistent with an $s$ orbital.
- $s = +\frac{1}{2}$ indicates the spin of the electron.

Step 3: The electronic configuration of the valence electron is $4s^1$ .

Step 4: Identify the element with this valence electron configuration. Potassium (K) has the electronic configuration [Ar] $4s^1$ .

Final Answer: K

Why other options are wrong:
- Option B (Ti): Titanium (Ti) has the electronic configuration [Ar] $4s^2 3d^2$ .
- Option C (Na): Sodium (Na) has the electronic configuration [Ne] $3s^1$ .
- Option D (Sc): Scandium (Sc) has the electronic configuration [Ar] $4s^2 3d^1$ .

Key Concepts:
- Quantum Numbers (n, l, ml, s)
- Electronic Configuration
- Valence Electrons

Memory Aids:
- Remember the relationship between $l$ and the orbital type: $l = 0$ (s), $l = 1$ (p), $l = 2$ (d), $l = 3$ (f).

Quick Tips:
- Use the quantum numbers to determine the orbital and then identify the element with that valence electron configuration.

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Application +4 | -1

H has two natural isotopes of

_1{\mathrm{H}^1}

and

_1{\mathrm{H}^2}

and O has two isotopes

\mathrm{\;}\,{O^{16}}

and

\mathrm{\;}\,{O^{18}}

. Which of the following mol.wt.of

{H_2}O

will not be possible?

View answer and explanation

Correct option: C

24

Option C (24) is correct because the maximum possible molecular weight of H2O is achieved with the heaviest isotopes of hydrogen and oxygen, which is less than 24.

Solution approach

Correct Answer + Core Reason: 24. The maximum molecular weight of H2O using the given isotopes is less than 24.

Step-by-Step Solution:

Step 1: Identify the isotopes of hydrogen and oxygen: $_1{\mathrm{H}^1}$ , $_1{\mathrm{H}^2}$ , ${\mathrm{O}^{16}}$ , ${\mathrm{O}^{18}}$ .

Step 2: Determine the possible molecular weights of H2O using these isotopes. The general formula is $2 \times (H \text{{ isotope mass}}) + (O \text{{ isotope mass}})$ .

Step 3: Calculate the possible molecular weights:
- $2(1) + 16 = 18$
- $2(1) + 18 = 20$
- $2(2) + 16 = 20$
- $2(2) + 18 = 22$

Step 4: Identify which molecular weight is NOT possible from the options.

Step 5: The possible molecular weights are 18, 20, and 22. 24 is not possible.

Why other options are wrong:
- Option A (19): This is possible with a combination of isotopes not listed, but not with the provided isotopes.
- Option B (20): This is a possible molecular weight using the given isotopes: $2(1) + 18 = 20$ or $2(2) + 16 = 20$ .
- Option D (22): This is a possible molecular weight using the given isotopes: $2(2) + 18 = 22$ .

Key Concepts:
- Isotopes and their masses
- Molecular weight calculation

Memory Aids:
- Molecular weight is the sum of atomic weights of all atoms in a molecule.

Quick Tips:
- Always consider all possible combinations of isotopes when calculating molecular weights.

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Application +4 | -1

Energy of H-atom in the ground state is -13.6 eV, hence energy in the second excited state is

View answer and explanation

Correct option: C

- 1.51\mathrm{eV}

Option C (-1.51 eV) is correct because the energy in the nth state is given by -13.6/n², and the second excited state corresponds to n=3.

Solution approach

Correct Answer + Core Reason: $- 1.51\mathrm{\;eV}$ . The energy of the H-atom in the nth state is given by $E_n = \frac{{-13.6}}{{n^2}} eV$ .

Step-by-Step Solution:

Step 1: Identify the given information: Energy of H-atom in the ground state, $E_1 = -13.6 eV$ .\nWe need to find the energy in the second excited state, which corresponds to $n=3$ .

Step 2: Use the formula for the energy of an electron in the nth orbit: $E_n = \frac{{-13.6}}{{n^2}} eV$

Step 3: Substitute the values: $E_3 = \frac{{-13.6}}{{3^2}} eV = \frac{{-13.6}}{9} eV$

Step 4: Calculate the energy: $E_3 = -1.51 eV$

Step 5: Final Answer: $E_3 = -1.51 eV$

Why other options are wrong:
- Option A: This is incorrect. It might be the energy of the electron in the second orbit (n=2).
- Option B: This is incorrect. It might be the energy of the electron in the second orbit (n=2) divided by 2.
- Option D: This is incorrect. There is no clear calculation that would lead to this answer.

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

Memory Aids:
- Remember the formula $E_n = \frac{{E_1}}{{n^2}}$

Quick Tips:
- The ground state corresponds to n=1, the first excited state to n=2, the second excited state to n=3, and so on.

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Analyzing +4 | -1

Which of the following has more number of unpaired electron?

View answer and explanation

Correct option: B

\mathrm{F}{\mathrm{e}^{2 + }}

Option B (Fe²⁺) is correct because it has the electronic configuration [Ar] 3d⁶, resulting in 4 unpaired electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}^{2 + }}$ . Fe²⁺ has the most unpaired electrons.

Step-by-Step Solution:

Step 1: Write the electronic configurations for each species.\nZn⁺: [Ar] 3d¹⁰ 4s¹ (1 unpaired electron)\nFe²⁺: [Ar] 3d⁶ (4 unpaired electrons)\nNi²⁺: [Ar] 3d⁸ (2 unpaired electrons)\nCu⁺: [Ar] 3d¹⁰ (0 unpaired electrons)

Step 2: Determine the number of unpaired electrons in each case.

Step 3: Compare the number of unpaired electrons.\nFe²⁺ has the highest number of unpaired electrons (4).

Why other options are wrong:
- Option A: Zn⁺ has only 1 unpaired electron, which is less than Fe²⁺.
- Option C: Ni²⁺ has only 2 unpaired electrons, which is less than Fe²⁺.
- Option D: Cu⁺ has 0 unpaired electrons, which is less than Fe²⁺.

Key Concepts:
- Electronic configuration
- Hund's rule
- Unpaired electrons

Memory Aids:
- Remember Hund's rule: electrons fill orbitals singly before pairing up.

Quick Tips:
- Write out the electronic configurations to easily determine the number of unpaired electrons.

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Application +4 | -1

A particular station of All India Radio, New Delhi broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is: [speed of light

\mathrm{c} = 3.0 \times {10^8}\mathrm{ m}{\mathrm{s}^{ - 1}}

]

View answer and explanation

Correct option: C

219.3 m

Option D (219.2 m) is the closest to the correct wavelength calculated using the formula wavelength = speed of light / frequency.

Solution approach

Correct Answer + Core Reason: 219.3 m. The wavelength is calculated using the formula: wavelength = speed of light / frequency.

Step-by-Step Solution:

Step 1: Identify the given information: frequency ( $\nu$ ) = 1368 kHz = 1368 × 10³ Hz, speed of light (c) = 3.0 × 10⁸ m/s.

Step 2: Recall the formula relating wavelength, frequency, and speed of light: $c = \nu \lambda$ , where $\lambda$ is the wavelength.

Step 3: Rearrange the formula to solve for wavelength: $\lambda = \frac{c}{\nu}$

Step 4: Substitute the given values into the formula: $\lambda = \frac{{3.0 \times 10^8 \text{{ m/s}}}}{{1368 \times 10^3 \text{{ Hz}}}}$

Step 5: Calculate the wavelength: $\lambda = \frac{{3.0 \times 10^8}}{{1368 \times 10^3}} \text{{ m}} = 219.29 \text{{ m}} \approx 219.3 \text{{ m}}$

Step 6: Final Answer: 219.3 m

Why other options are wrong:
- Option A (2192 m): This is incorrect because the frequency was not properly converted from kHz to Hz before calculation, or the division was done incorrectly.
- Option B (21.92 cm): This is incorrect because the units are wrong and the calculation is off by several orders of magnitude. It may be a result of incorrectly converting meters to centimeters after the calculation.
- Option C (219.3 m): This is the correct answer.

Key Concepts:
- Relationship between speed of light, frequency, and wavelength: $c = \nu \lambda$
- Unit conversions (kHz to Hz)

Memory Aids:
- Remember the formula: c = $\lambda$ $\nu$ (speed of light equals wavelength times frequency)

Quick Tips:
- Always convert all units to SI units before performing calculations.
- Double-check your calculations to avoid errors.

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Remembering +4 | -1

The work-function for photoelectric effect :

View answer and explanation

Correct option: C

Is different for different metals

Option C (Is different for different metals) is correct because the work function is a material property that depends on the metal's electronic structure.

Solution approach

Correct Answer + Core Reason: Is different for different metals. The work function is a material property.

Step-by-Step Solution:

Step 1: Define the work function.

Step 2: The work function ( $\phi$ ) is the minimum energy required to remove an electron from the surface of a solid.

Step 3: The work function depends on the type of metal because different metals have different electronic structures and binding energies for their electrons.

Step 4: Therefore, the work function is different for different metals.

Why other options are wrong:
- Option A: The work function is a property of the metal and does not depend on the frequency of incident light. The kinetic energy of the emitted electrons depends on the frequency, but not the work function itself.
- Option B: The work function is not the same for all metals, as explained above.
- Option D: The work function exists and is different for different metals, so this option is incorrect.

Key Concepts:
- Photoelectric effect
- Work function
- Material properties

Memory Aids:
- Remember that the work function is a characteristic property of a metal.

Quick Tips:
- Understand the definition of work function and its relation to the photoelectric effect.

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Understanding +4 | -1

A hydrogen atom in its ground state absorbs a photon. The maximum energy of such a photon is:

View answer and explanation

Correct option: D

13.6 eV

Option D (13.6 eV) is correct because the maximum energy a ground state hydrogen atom can absorb corresponds to its ionization energy, which is 13.6 eV.

Solution approach

Correct Answer + Core Reason: 13.6 eV. The maximum energy a ground state hydrogen atom can absorb is its ionization energy.

Step-by-Step Solution:

Step 1: The question asks for the maximum energy a photon can have to be absorbed by a hydrogen atom in its ground state.

Step 2: Recall that the ground state energy of a hydrogen atom is -13.6 eV. The maximum energy that can be absorbed corresponds to the ionization energy, which is the energy required to remove the electron completely from the atom.

Step 3: The ionization energy is the energy difference between the ground state (-13.6 eV) and the state where the electron is completely removed (0 eV). Therefore, the maximum energy is 0 - (-13.6 eV) = 13.6 eV.

Why other options are wrong:
- Option A: 1.5 eV is incorrect because it is a much lower energy than the ionization energy.
- Option B: 3.4 eV is incorrect because it corresponds to the energy difference between n=1 and n=2, not the ionization energy.
- Option C: 10.2 eV is incorrect because it corresponds to the energy difference between n=1 and n=3, not the ionization energy.

Key Concepts:
- Ionization energy
- Ground state energy
- Hydrogen atom energy levels

Memory Aids:
- Ionization energy of hydrogen = 13.6 eV

Quick Tips:
- Remember the ionization energy of hydrogen.

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Remembering +4 | -1

Gamma rays are:

View answer and explanation

Correct option: C

High energy electromagnetic waves

Option C (High energy electromagnetic waves) is correct because gamma rays are a form of electromagnetic radiation with very high energy and short wavelengths.

Solution approach

Correct Answer + Core Reason: High energy electromagnetic waves. Gamma rays are part of the electromagnetic spectrum with the highest energy.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum.

Step 2: Identify the position of gamma rays in the spectrum. Gamma rays are at the highest energy end, beyond X-rays.

Step 3: Relate energy to the nature of the wave. High energy corresponds to high frequency and short wavelength, characteristic of electromagnetic waves.

Step 4: Conclude that gamma rays are high-energy electromagnetic waves.

Why other options are wrong:
- Option A: High energy electrons are beta particles, not gamma rays.
- Option B: Low energy electrons are not related to gamma rays.
- Option D: High energy positrons are antimatter counterparts of electrons, not gamma rays.

Key Concepts:
- Electromagnetic spectrum
- Gamma rays
- Energy of electromagnetic radiation

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma (from low to high energy).

Quick Tips:
- Understand the different types of radiation and their properties.

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Application +4 | -1

The wavelength of radiation emitted when electron falls from 4th Bohr's orbit to 2nd in H-atom is:

\left( {{R_\mathrm{H}} = - 1.09678 \times {{10}^{ - 7}}{\mathrm{m}^{ - 1}}} \right)

View answer and explanation

Correct option: B

486 nm

Option B (486 nm) is correct because it is calculated using the Rydberg formula for the wavelength of emitted radiation.

Solution approach

Correct Answer + Core Reason: 486 nm. The wavelength is calculated using the Rydberg formula.

Step-by-Step Solution:

Step 1: Identify the given values: $n_1 = 2$ , $n_2 = 4$ , and $R_H = 1.09678 \times 10^7 m^{-1}$ .

Step 2: Use the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ .

Step 3: Substitute the given values into the formula: $\frac{1}{\lambda} = 1.09678 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$ .

Step 4: Simplify the expression: $\frac{1}{\lambda} = 1.09678 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.09678 \times 10^7 \left( \frac{4 - 1}{16} \right) = 1.09678 \times 10^7 \left( \frac{3}{16} \right)$ .

Step 5: Calculate $\frac{1}{\lambda} = 0.20563125 \times 10^7 m^{-1}$ .

Step 6: Find the wavelength: $\lambda = \frac{1}{0.20563125 \times 10^7} = 4.862 \times 10^{-7} m = 486.2 nm$ .

Step 7: Round the wavelength to 486 nm.

Why other options are wrong:
- Option A: This is incorrect and likely results from an error in applying the Rydberg formula or incorrect arithmetic.
- Option C: This is incorrect and likely results from an error in applying the Rydberg formula or incorrect arithmetic.
- Option D: This is incorrect and likely results from an error in applying the Rydberg formula or incorrect arithmetic.

Key Concepts:
- Rydberg formula
- Hydrogen spectrum
- Bohr's model

Memory Aids:
- Remember the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Quick Tips:
- Be careful with the units and ensure that the Rydberg constant is used with the correct units. Pay attention to the initial and final energy levels.

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Remembering +4 | -1

Bohr frequency condition is given by

View answer and explanation

Correct option: A

v = \frac{{\Delta E}}{h}

Option A (

v = \frac{{\Delta E}}{h}

) is correct because the Bohr frequency condition relates the frequency of emitted radiation to the energy difference between energy levels and Planck's constant.

Solution approach

Correct Answer + Core Reason: $v = \frac{{\Delta E}}{h}$ . The Bohr frequency condition relates the frequency of emitted radiation to the energy difference between energy levels and Planck's constant.

Step-by-Step Solution:

Step 1: Bohr's Frequency Condition: According to Bohr's model, when an electron transitions from a higher energy level ( $E_2$ ) to a lower energy level ( $E_1$ ), a photon is emitted with energy equal to the energy difference between the levels:
$\Delta E = E_2 - E_1 = h \times v$ , where $h$ is Planck's constant and $v$ is the frequency of the emitted photon.

Step 2: Solving for Frequency: Rearranging the equation, we get the Bohr frequency condition:\nv = $\frac{{Δ E}}{h}$ $\n\n**Why other options are wrong:**\n- **Option B:**$ v = $\frac{{Δ E}}{{hc}}$ $. This is incorrect. It would give units of inverse wavelength, not frequency.\n- **Option C:**$ v = $\frac{{hc}}{{Δ E}}$ $. This is incorrect. It represents the inverse of the correct relationship.\n- **Option D:**$ v = $\frac{h}{{Δ E}}$ $. This is incorrect. It represents the inverse of the correct relationship and has incorrect units.\n\n**Key Concepts:**\n- Bohr's model of the hydrogen atom\n- Bohr frequency condition\n- Energy levels\n- Planck's constant\n\n**Memory Aids:**\n-$ Δ E = h × v $(Energy difference equals Planck's constant times frequency).\n\n**Quick Tips:**\n- Remember that the frequency of emitted radiation is directly proportional to the energy difference between energy levels.$

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Application +4 | -1

Photons of energy 6 eV are incidented on a potassium surface of work function 2.1 eV. What is the stopping potential?

View answer and explanation

Correct option: C

- 3.9\mathrm{\;V}

Option C ( - 3.9 V) is correct because the stopping potential is equal to the maximum kinetic energy of the emitted electrons divided by the elementary charge, and the maximum kinetic energy is the difference between the photon energy and the work function.

Solution approach

Correct Answer + Core Reason: $- 3.9\mathrm{\;V}$ . The stopping potential is the negative of the maximum kinetic energy divided by the electron charge.

Step-by-Step Solution:

Step 1: The energy of the incident photons is 6 eV, and the work function of the potassium surface is 2.1 eV.

Step 2: The maximum kinetic energy ( $KE_{max}$ ) of the emitted photoelectrons is given by:
$KE_{max} = E_{photon} - W$ \nwhere $E_{photon}$ is the energy of the incident photon and W is the work function.

Step 3: Substitute the given values:
$KE_{max} = 6 \text{ eV} - 2.1 \text{ eV} = 3.9 \text{ eV}$

Step 4: The stopping potential ( $V_s$ ) is the potential required to stop the most energetic photoelectrons. It is related to the maximum kinetic energy by:
$KE_{max} = e |V_s|$ \nwhere e is the elementary charge.

Step 5: Therefore, the stopping potential is:
$|V_s| = \frac{KE_{max}}{e} = \frac{3.9 \text{ eV}}{e} = 3.9 \text{ V}$ \nSince the stopping potential is negative, $V_s = -3.9 \text{ V}$

Why other options are wrong:
- Option A: Incorrect. This is the energy of the incident photons, not the stopping potential.
- Option B: Incorrect. This is the work function, not the stopping potential.
- Option D: Incorrect calculation.

Key Concepts:
- Photoelectric effect
- Work function
- Maximum kinetic energy of photoelectrons
- Stopping potential

Memory Aids:
- $KE_{max} = hf - W$
- $KE_{max} = e |V_s|$

Quick Tips:
- Remember the relationship between the stopping potential and the maximum kinetic energy of the photoelectrons.

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