CUET UG Physics Booster Test 3-Electron Emission and Photoelectric Observations -

🚀

CUET UG Physics Booster Test 3-Electron Emission and Photoelectric Observations

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3-Electron Emission and Photoelectric Observations

1 / 20

QUESTION 1 OF 20

If an electron in a metal requires exactly 3.428 × 10⁻¹⁹ J to overcome the attractive pull and escape the surface, what is the work function of this metal in electron volts (eV)? (1 eV = 1.602 × 10⁻¹⁹ J)

Answer Review

1 If an electron in a metal requires exactly 3.428 × 10⁻¹⁹ J to overcome the attractive pull and escape the surface, what is the work function of this metal in electron volts (eV)? (1 eV = 1.602 × 10⁻¹⁹ J)

A 1.60 eV

B 2.14 eV

C 3.42 eV

D 5.48 eV

💡 Short Explanation

�� Work function can be expressed in eV. �� Convert Joules into eV using the given conversion factor. �� Divide energy in Joules by 1.602 × 10⁻¹⁹.

✅ Correct Answer Explanation

Given: Work function = 3.428 × 10⁻¹⁹ J 1 eV = 1.602 × 10⁻¹⁹ J → Therefore, Work function = (3.428 × 10⁻¹⁹)/(1.602 × 10⁻¹⁹) = 2.14 eV → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Smaller than the calculated value.
�� Option C → Incorrect conversion.
�� Option D → Much larger than the actual value.

🎯 Strategy

Used

Substitution

Application:

Substitute the given conversion factor and calculate.

Final Logic:

3.428 × 10⁻¹⁹ J corresponds to 2.14 eV.

🧠 Memory Trick

J → eV ⇒ Divide by 1.602 × 10⁻¹⁹

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Work Function and Electron Volt

2 Match List I (Parameters) with List II (Influence on Work Function φ₀)

List I	List II (Correct Match)
(1) Nature of the surface	(a) Strongly determines φ₀ alongside the metal properties
(2) Magnetic field of the surrounding air	(b) Has no effect on the inherent φ₀
(3) Properties of the metal	(c) Strongly determines φ₀ alongside the surface nature
(4) Frequency of incident visible light	(d) Irrelevant to defining the material's threshold φ₀

A (1)-d, (2)-c, (3)-a, (4)-b

B (1)-c, (2)-b, (3)-a, (4)-d

C (1)-a, (2)-b, (3)-c, (4)-d

D (1)-b, (2)-d, (3)-c, (4)-a

💡 Short Explanation

�� Work function depends on metal and surface. �� Magnetic field does not determine work function. �� Light frequency does not define the inherent work function.

✅ Correct Answer Explanation

The correct matching is: → (1) Nature of the surface → (a) Strongly determines φ₀ alongside the metal properties → (2) Magnetic field of the surrounding air → (b) Has no effect on the inherent φ₀ → (3) Properties of the metal → (c) Strongly determines φ₀ alongside the surface nature → (4) Frequency of incident visible light → (d) Irrelevant to defining the material's threshold φ₀ → Work function is an intrinsic property that depends mainly on the metal and surface condition.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly assigns magnetic field effects.
�� Option B → Incorrectly interchanges surface nature and metal properties.
�� Option D → Incorrectly assigns dependence on incident light frequency.

🎯 Strategy

Used

Option Grouping

Application:

Match each factor with its actual influence on work function.

Final Logic:

Only Option A correctly associates all parameters.

🧠 Memory Trick

Metal + Surface = Work Function

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Work Function

3 The kinetic energy K gained by an electron accelerated from rest through a potential V can be expressed mathematically as the product of its charge and potential. If e is the elementary charge, K is:

A e / V

B eV

C V / e

D e²V

💡 Short Explanation

�� Energy gained equals charge × potential difference. �� Electron charge is e. �� Therefore K = eV.

✅ Correct Answer Explanation

When a charge moves through a potential difference V, the electrical energy gained is: K = qV → For an electron, K = eV → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect relationship.
�� Option C → Not an energy expression.
�� Option D → Contains an extra factor of e.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Apply the basic formula K = qV.

Final Logic:

The kinetic energy gained by an electron is eV.

🧠 Memory Trick

Energy = Charge × Voltage

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Electron Volt

4 The conversion factor between Joules and electron volts highlights the scale of atomic physics. Which statement correctly interprets this relationship based on the text?

A 1 Joule is equivalent to 1.602 × 10⁻¹⁹ eV, making Joules the microscopic unit.

B The macroscopic nature of Joules makes it 1.602 × 10¹⁹ times smaller than an eV.

C An electron volt is a microscopic energy unit equal to 1.602 × 10⁻¹⁹ Joules, mapping directly to the energy gained by accelerating one electron through one volt.

D 1 eV represents the power output of 1 Joule per second by an electron.

💡 Short Explanation

�� eV is a unit of energy. �� 1 eV = 1.602 × 10⁻¹⁹ J. �� Widely used in atomic and nuclear physics.

✅ Correct Answer Explanation

One electron volt is the energy gained by an electron accelerated through a potential difference of one volt. → Numerically, 1 eV = 1.602 × 10⁻¹⁹ J → Therefore Option C correctly interprets the physical meaning of the conversion.

❓ Why Other Options Are Incorrect

�� Option A → The conversion is reversed.
�� Option B → Joule is much larger, not smaller.
�� Option D → eV is a unit of energy, not power.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the statement matching the standard definition of eV.

Final Logic:

Option C accurately defines electron volt and its significance.

🧠 Memory Trick

1 Electron × 1 Volt = 1 eV

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Electron Volt

5 Statements regarding thermionic emission
1. It provides energy to electrons by suitably heating the metal.
2. The supplied energy allows electrons to come out of the metal.
3. It relies on high-frequency light rather than thermal heat.
4. It imparts thermal energy to the free electrons.

A (1), (2), (3)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Thermionic emission is heat based. �� Heating supplies thermal energy. �� High-frequency light is not involved.

✅ Correct Answer Explanation

Statement 1 is correct because the metal is suitably heated. → Statement 2 is correct because the supplied thermal energy allows electrons to escape. → Statement 3 is incorrect because high-frequency light is associated with photoelectric emission. → Statement 4 is correct because thermal energy is imparted to free electrons. → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 3.
�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Check each statement using the definition of thermionic emission.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Thermionic = Thermal Energy

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Thermionic Emission

6 Correct statements about the mechanics of thermionic emission
1. Suitably heating imparts thermal energy to free electrons.
2. The attractive forces of the ions must be overcome by this thermal energy.
3. It requires an electric field of the order of 10⁸ V m⁻¹.
4. It is one of the physical processes to supply the minimum escape energy.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (3)

💡 Short Explanation

�� Heating gives energy to electrons. �� Electrons overcome ionic attraction. �� Strong electric fields belong to field emission.

✅ Correct Answer Explanation

Statement 1 is correct because heating supplies thermal energy. → Statement 2 is correct because electrons must overcome the attractive force of ions. → Statement 3 is incorrect because 10⁸ V m⁻¹ is associated with field emission. → Statement 4 is correct because thermionic emission is one method of supplying escape energy. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Separate thermionic emission concepts from field emission concepts.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Heat Beats Ion Attraction

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Thermionic Emission

7 Field emission is physically able to extract electrons out of a metal because

A The applied magnetic field repels the electrons from the interior

B A very strong electric field of the order of 10⁸ V m⁻¹ pulls the electrons out

C The high voltage generates extreme localized heating, causing thermionic emission

D The external electric field lowers the work function to exactly zero

💡 Short Explanation

�� Strong electric fields extract electrons. �� Typical field strength is about 10⁸ V m⁻¹. �� Heating is not the primary mechanism.

✅ Correct Answer Explanation

In field emission, an extremely strong electric field acts on electrons near the surface. → The field pulls electrons out of the metal. → Typical field strength is of the order of 10⁸ V m⁻¹. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnetic fields do not produce field emission.
�� Option C → Describes thermionic emission.
�� Option D → The work function does not become exactly zero.

🎯 Strategy

Used

Elimination

Application:

Identify the mechanism unique to field emission.

Final Logic:

Electron extraction occurs due to a strong electric field.

🧠 Memory Trick

Field Emission = 10⁸ V m⁻¹

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Field Emission

8 Incorrect statement about the application of electric field emission

A It applies to everyday mechanical devices like spark plugs.

B It requires a strong external electric field (10⁸ V m⁻¹).

C The process strictly relies on irradiating the metal with ultraviolet light.

D Electrons are pulled directly out of the metal by the field.

💡 Short Explanation

�� Field emission requires strong electric fields. �� UV radiation is associated with photoelectric emission. �� Electrons are directly extracted by the field.

✅ Correct Answer Explanation

Electric field emission occurs due to the action of a strong electric field. → Ultraviolet light is not required for field emission. → UV radiation is associated with photoelectric emission. → Therefore Option C is the correct statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct application.
�� Option B → Correct field strength requirement.
�� Option D → Correct description of the process.

🎯 Strategy

Used

Odd One Out

Application:

Identify the option describing a different emission mechanism.

Final Logic:

UV radiation belongs to photoelectric emission, not field emission.

🧠 Memory Trick

Field = Electric, Photo = Light

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Field Emission

9 In the photoelectric effect, electrons near the surface absorb enough energy from the ________ to overcome the ________ of the positive ions.

A Electric field, Repulsion

B Incident radiation, Attraction

C Thermal heat, Attraction

D Magnetic field, Repulsion

💡 Short Explanation

�� Incident radiation supplies energy. �� Positive ions attract electrons. �� Electrons escape after overcoming attraction.

✅ Correct Answer Explanation

During the photoelectric effect, electrons absorb energy from incident electromagnetic radiation. → This energy enables them to overcome the attractive force exerted by the positive ions. → Therefore the correct combination is Incident radiation and Attraction.

❓ Why Other Options Are Incorrect

�� Option A → Positive ions attract rather than repel electrons.
�� Option C → Describes thermionic emission.
�� Option D → Magnetic fields are not responsible.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the photoelectric process with the correct physical concepts.

Final Logic:

Incident radiation provides energy to overcome ionic attraction.

🧠 Memory Trick

Light In → Electron Out

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photoelectric Effect

10 The fundamental identity of a photoelectron

A is identical to any other electron, with the term "photoelectrons" merely indicating its light-induced generation

B differs completely from a cathode ray particle due to its positive mass

C possesses a completely different charge-to-mass (e/m) ratio compared to thermionic electrons

D indicates it is a photon masquerading as a charged particle

💡 Short Explanation

�� A photoelectron is an ordinary electron. �� Only the mode of production differs. �� Charge and mass remain unchanged.

✅ Correct Answer Explanation

Photoelectrons are ordinary electrons emitted due to incident light. → Their charge, mass and e/m ratio are identical to all other electrons. → The term "photoelectron" only indicates how the electron was produced. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Electrons do not have positive mass.
�� Option C → All electrons have the same e/m ratio.
�� Option D → A photoelectron is not a photon.

🎯 Strategy

Used

Elimination

Application:

Compare the known properties of electrons.

Final Logic:

A photoelectron is simply an electron emitted through the photoelectric effect.

🧠 Memory Trick

Photoelectron = Electron + Light Origin

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photoelectrons and Photoelectric Effect

11 Hertz's 1887 observations linked electromagnetic wave generation with photoelectric emission. What was the critical detail of his observation?

A Producing sparks caused the arc lamp to emit UV light.

B Illuminating the detector with IR light completely suppressed the spark discharge.

C Illuminating the emitter plate with UV light facilitated free electron escape, enhancing the spark discharge.

D Heating the detector loop caused it to emit UV light and enhance sparks.

💡 Short Explanation

�� Hertz used spark discharges to study electromagnetic waves. �� Ultraviolet light enhanced spark discharge. �� Electron emission from the metal surface caused the enhancement.

✅ Correct Answer Explanation

Hertz observed that when ultraviolet light from an arc lamp illuminated the emitter plate, spark discharge became stronger. → The ultraviolet radiation helped electrons escape from the metal surface. → This facilitated electrical discharge and enhanced the spark across the detector loop. → Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The arc lamp produced UV light; sparks did not cause the arc lamp to emit UV.
�� Option B → Infrared light was not responsible for the observed enhancement.
�� Option D → Heating the detector loop was not the mechanism observed.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the historical observation with the actual experimental result.

Final Logic:

UV-assisted electron emission enhanced spark discharge.

🧠 Memory Trick

UV on Metal → Stronger Spark

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Hertz's Observations

12 Statements concerning the impact of ultraviolet light in Hertz's setup
1. Ultraviolet light was sourced from an arc lamp.
2. It illuminated the detector loop directly.
3. It illuminated the emitter plate.
4. It caused the spark discharge across the detector loop to be enhanced.

A (1), (3), (4)

B (1), (2), (3)

C (2), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Arc lamp provided ultraviolet light. �� UV light illuminated the emitter plate. �� Spark discharge became stronger.

✅ Correct Answer Explanation

Statement 1 is correct because the UV radiation came from an arc lamp. → Statement 2 is incorrect because the significant illumination was on the emitter plate, not directly on the detector loop. → Statement 3 is correct because the emitter plate was illuminated. → Statement 4 is correct because the spark discharge became enhanced. → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 2.
�� Option C → Includes incorrect statement 2.
�� Option D → Includes incorrect statement 2.

🎯 Strategy

Used

Option Grouping

Application:

Verify each statement using Hertz's experimental observations.

Final Logic:

Only statements 1, 3 and 4 are correct.

🧠 Memory Trick

Arc Lamp → UV → Emitter → Spark

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Hertz's Experiment

13 Correct statements derived from Hallwachs' zinc plate experiments
1. A negatively charged zinc plate lost its charge under UV irradiation.
2. The loss of charge logically indicated the emission of negative particles.
3. Positive charge on a positively charged zinc plate was further enhanced.
4. UV light had absolutely no effect on an uncharged zinc plate.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (3)

💡 Short Explanation

�� Negative zinc plates lost charge. �� Emitted particles were negatively charged. �� Positive charge increased on positively charged plates.

✅ Correct Answer Explanation

Statement 1 is correct because UV light caused the negatively charged zinc plate to discharge. → Statement 2 is correct because the loss of negative charge implied emission of negatively charged particles (electrons). → Statement 3 is correct because a positively charged zinc plate became even more positively charged when electrons were emitted. → Statement 4 is incorrect because an uncharged zinc plate became positively charged under UV illumination. → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 4.
�� Option B → Includes incorrect statement 4.
�� Option C → Includes incorrect statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Check each observation against Hallwachs' experimental results.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

UV → Electrons Leave → Positive Charge Increases

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Hallwachs Zinc Plate Experiment

14 Incorrect statement regarding Hallwachs' observations of the positively charged zinc plate

A It became further positively charged when illuminated by UV light.

B The enhancement of positive charge implied the emission of positively charged ions from the plate.

C The observation supported the idea that negatively charged particles were leaving the plate.

D He utilized an electroscope to monitor the charge.

💡 Short Explanation

�� Positive charge increased due to electron loss. �� Electrons are negatively charged. �� Hallwachs used an electroscope.

✅ Correct Answer Explanation

The positively charged zinc plate became more positively charged because electrons left the surface. → Emission of positively charged ions would not explain the photoelectric effect observed. → The experiment supported the emission of negatively charged electrons. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct observation.
�� Option C → Correct interpretation of electron emission.
�� Option D → Hallwachs used an electroscope to detect charge changes.

🎯 Strategy

Used

Elimination

Application:

Determine which statement contradicts the electron-emission interpretation.

Final Logic:

The emitted particles were electrons, not positive ions.

🧠 Memory Trick

More Positive = Electrons Lost

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Hallwachs Experiment

15 In Lenard's setup, the potential difference between the positive collector plate A and emitter plate C is V. If an electron travels from C to A, the work done by the electric field on the electron is:

A eV

B -eV

C e / V

D V / e

💡 Short Explanation

�� Electric field accelerates the electron. �� Work done equals charge × potential difference. �� Energy gained is eV.

✅ Correct Answer Explanation

An electron moving through a potential difference V gains kinetic energy equal to: Work Done = eV → Here e is the magnitude of electronic charge. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect sign for energy gained.
�� Option C → Incorrect expression.
�� Option D → Not a work expression.

🎯 Strategy

Used

Substitution

Application:

Use Work Done = Charge × Potential Difference.

Final Logic:

Electron gains energy equal to eV.

🧠 Memory Trick

Energy Gain = e × V

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photoelectric Current and Potential Difference

16 Match the action in Lenard's experiment (List I) with its consequence on current flow (List II).

List I	List II
(1) UV radiations fall on the emitter plate	(a) Results in current flow in the external circuit
(2) UV radiations are abruptly stopped	(b) Electrons are ejected from the emitter plate C
(3) Electrons are attracted towards the positive collector A	(c) Current flow in the circuit immediately stops
(4) Electrons flow through the evacuated tube	(d) The flow across the gap results in the circuit current

A (1)-b, (2)-c, (3)-d, (4)-a

B (1)-a, (2)-b, (3)-c, (4)-d

C (1)-c, (2)-d, (3)-a, (4)-b

D (1)-d, (2)-a, (3)-b, (4)-c

💡 Short Explanation

�� UV light ejects electrons. �� Removing UV stops emission. �� Electron motion produces current.

✅ Correct Answer Explanation

The correct matching is: → (1) UV radiations fall on the emitter plate → (b) Electrons are ejected from the emitter plate C → (2) UV radiations are abruptly stopped → (c) Current flow in the circuit immediately stops → (3) Electrons are attracted towards the positive collector A → (d) The flow across the gap results in the circuit current → (4) Electrons flow through the evacuated tube → (a) Results in current flow in the external circuit → This sequence describes the operation of Lenard's photoelectric tube.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly matches UV illumination and current generation.
�� Option C → Multiple action-consequence pairs are incorrect.
�� Option D → Incorrectly reverses several relationships.

🎯 Strategy

Used

Option Grouping

Application:

Match each experimental event with its direct consequence.

Final Logic:

Only Option A correctly follows the photoelectric process.

🧠 Memory Trick

UV → Electrons → Current

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Lenard's Experiment

17 The ________ frequency is a sharp boundary; if the incident light frequency is below it, photoelectric emission is absolutely ________.

A Maximum, Guaranteed

B Threshold, Impossible

C Cut-off, Instantaneous

D Saturation, Impossible

💡 Short Explanation

�� Threshold frequency is the minimum required frequency. �� Below threshold, emission never occurs. �� Intensity cannot overcome this limit.

✅ Correct Answer Explanation

Threshold frequency is the minimum frequency required to produce photoelectric emission. → If the incident frequency is below this value, electrons are not emitted regardless of intensity. → Therefore the correct combination is Threshold and Impossible.

❓ Why Other Options Are Incorrect

�� Option A → Opposite of the actual concept.
�� Option C → Instantaneous does not describe the condition.
�� Option D → Saturation frequency is not a standard term.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the definition of threshold frequency.

Final Logic:

Below threshold frequency, emission is impossible.

🧠 Memory Trick

Below Threshold = No Emission

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Threshold Frequency

18 The observation that the threshold frequency strictly depends on the nature of the material of the emitter plate implies that

A Different materials have identically sized atomic nuclei

B Glass tubes alter the frequency of the incoming light based on the metal inside

C The intensity of light can compensate for a higher threshold

D Different metals possess varying inherent attractive forces holding their electrons

💡 Short Explanation

�� Different metals have different work functions. �� Therefore threshold frequencies differ. �� Electron binding strength varies among materials.

✅ Correct Answer Explanation

Threshold frequency depends on the work function of the material. → Different metals hold electrons with different strengths. → Hence different materials possess different threshold frequencies. → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Nuclear size is unrelated.
�� Option B → Glass tubes do not alter frequency in this manner.
�� Option C → Intensity cannot compensate for insufficient frequency.

🎯 Strategy

Used

Elimination

Application:

Relate threshold frequency to work function.

Final Logic:

Different electron binding strengths produce different threshold frequencies.

🧠 Memory Trick

Higher Binding → Higher Threshold

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Threshold Frequency and Work Function

19 Based on their photoelectric sensitivity profiles, what primarily differentiates alkali metals (like Li, Na, K) from other metals (like Zn, Cd)?

A Alkali metals have much higher threshold frequencies and need X-rays.

B Alkali metals are sensitive to visible light, while zinc and cadmium respond only to shorter-wavelength ultraviolet light.

C Heavy metals emit protons, while alkali metals emit electrons.

D Alkali metals only emit electrons when heated via thermionic emission, not when illuminated.

💡 Short Explanation

�� Alkali metals have low work functions. �� Visible light can eject electrons. �� Zinc and cadmium generally require UV radiation.

✅ Correct Answer Explanation

Alkali metals such as lithium, sodium and potassium are highly photosensitive. → Their lower work functions allow visible light to eject electrons. → Zinc and cadmium require higher-energy ultraviolet radiation. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Alkali metals require lower, not higher, threshold frequencies.
�� Option C → Both types emit electrons.
�� Option D → Alkali metals also show photoelectric emission.

🎯 Strategy

Used

Elimination

Application:

Compare the work functions and sensitivity of the metals.

Final Logic:

Lower work functions make alkali metals responsive to visible light.

🧠 Memory Trick

Alkali Metals Love Visible Light

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photosensitive Materials

20 If visible light has a lower frequency than ultraviolet light, and sodium emits electrons under visible light while zinc requires UV light, which metal must have a higher threshold frequency?

A Zinc

B Sodium

C Both have equal threshold frequencies

D Cannot be determined from the given text

💡 Short Explanation

�� Higher threshold frequency means higher energy requirement. �� Zinc requires ultraviolet light. �� Sodium responds to visible light.

✅ Correct Answer Explanation

Threshold frequency is the minimum frequency needed for photoelectric emission. → Since sodium emits electrons under visible light, its threshold frequency is relatively low. → Zinc requires ultraviolet light, which has a higher frequency than visible light. → Therefore zinc must have the higher threshold frequency. → Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Sodium has a lower threshold frequency.
�� Option C → Their threshold frequencies are not equal.
�� Option D → The information is sufficient to determine the answer.

🎯 Strategy

Used

Elimination

Application:

Compare the minimum frequencies required for emission.

Final Logic:

The metal requiring UV light must possess the higher threshold frequency.

🧠 Memory Trick

Needs UV → Higher Threshold

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Threshold Frequency and Photosensitive Materials

CUET UG Physics Booster Test 3-Electron Emission and Photoelectric Observations

Test Complete!

Answer Review