CUET UG Physics Booster Test 2-Electron Emission and Photoelectric Observations -

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CUET UG Physics Booster Test 2-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 2-Electron Emission and Photoelectric Observations

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QUESTION 1 OF 20

Match List I (Physical scenario) with List II (Inference related to escape energy)

List I	List II (Correct Match)
(1) Electron tries to leave metal	(a) Work function (φ₀)
(2) Metal surface post-electron attempt	(b) Acquires positive charge
(3) Retaining force inside metal	(c) Attractive forces of ions
(4) Minimum required escape energy	(d) Needs sufficient energy to overcome pull

Answer Review

1 Match List I (Physical scenario) with List II (Inference related to escape energy)

List I	List II (Correct Match)
(1) Electron tries to leave metal	(a) Work function (φ₀)
(2) Metal surface post-electron attempt	(b) Acquires positive charge
(3) Retaining force inside metal	(c) Attractive forces of ions
(4) Minimum required escape energy	(d) Needs sufficient energy to overcome pull

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

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

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

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

💡 Short Explanation

�� Electrons require sufficient energy to escape. �� Loss of electrons leaves the metal positively charged. �� Work function represents the minimum escape energy.

✅ Correct Answer Explanation

The correct matching is: → (1) Electron tries to leave metal → (d) Needs sufficient energy to overcome pull → (2) Metal surface post-electron attempt → (b) Acquires positive charge → (3) Retaining force inside metal → (c) Attractive forces of ions → (4) Minimum required escape energy → (a) Work function (φ₀) → The attractive ionic forces hold electrons within the metal, and the work function is the minimum energy needed to overcome these forces.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly matches electron escape and work function directly.
�� Option C → Incorrectly matches metal surface and retaining force.
�� Option D → Incorrectly matches electron escape and positive charge.

🎯 Strategy

Used

Option Grouping

Application:

Match each physical situation with its corresponding concept.

Final Logic:

Only Option A correctly relates all four concepts.

🧠 Memory Trick

Escape → Energy → Work Function

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

2 Statements about the work function's dependency
1. It changes based on the volume of the metal block.
2. It depends on the properties of the metal.
3. It depends on the nature of the surface.
4. It is generally denoted by φ₀ and measured in eV.

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

B (2), (3), (4)

C (1), (3), (4)

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

💡 Short Explanation

�� Work function is a material property. �� Surface condition affects work function. �� It is commonly expressed in electron volts.

✅ Correct Answer Explanation

Statement 1 is incorrect because work function does not depend on the volume of the metal. → Statement 2 is correct because work function depends on the nature and properties of the metal. → Statement 3 is correct because surface conditions influence work function. → Statement 4 is correct because work function is represented by φ₀ (or φ) and is commonly measured in eV. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 1.
�� Option C → Includes incorrect statement 1.
�� Option D → Includes incorrect statement 1.

🎯 Strategy

Used

Option Grouping

Application:

Verify each numbered statement individually.

Final Logic:

Only statements 2, 3 and 4 are correct.

🧠 Memory Trick

Metal + Surface = Work Function

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

3 If an electron (charge e) is accelerated by a potential difference of V volts, the energy gained is:

A e / V

B V / e

C eV

D eV²

💡 Short Explanation

�� Electrical energy equals charge × potential difference. �� Electron charge is e. �� Energy gained becomes eV.

✅ Correct Answer Explanation

The energy gained by a charge moving through a potential difference is given by: Energy = Charge × Potential Difference → Therefore, Energy = e × V = eV → Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect dimensional expression.
�� Option B → Does not represent energy.
�� Option D → Contains an extra factor of V.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use the formula Energy = qV.

Final Logic:

Energy gained by an electron through V volts is eV.

🧠 Memory Trick

Energy = Charge × Voltage

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

4 If the work function of a metal is 2.14 eV, what is this equivalent to in Joules? (1 eV = 1.602 × 10⁻¹⁹ J)

A 3.428 × 10⁻¹⁹ J

B 3.428 × 10⁻²⁰ J

C 2.140 × 10⁻¹⁹ J

D 1.602 × 10⁻¹⁹ J

💡 Short Explanation

�� Convert eV to Joules. �� Multiply by 1.602 × 10⁻¹⁹. �� Work function is an energy quantity.

✅ Correct Answer Explanation

Given: Work Function = 2.14 eV 1 eV = 1.602 × 10⁻¹⁹ J → Therefore, 2.14 × 1.602 × 10⁻¹⁹ = 3.428 × 10⁻¹⁹ J → Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → One power of ten too small.
�� Option C → Corresponds approximately to 1.34 eV.
�� Option D → Equivalent to only 1 eV.

🎯 Strategy

Used

Substitution

Application:

Directly substitute the conversion factor.

Final Logic:

2.14 eV converts to 3.428 × 10⁻¹⁹ J.

🧠 Memory Trick

eV × 1.602 × 10⁻¹⁹

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

5 Correct statements about thermionic emission
1. It is the process of extracting electrons using strong magnetic fields.
2. It is a physical process to supply minimum energy for electron emission.
3. It involves suitably heating the material.
4. It imparts thermal energy to free electrons.

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

B (2), (3), (4)

C (1), (3), (4)

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

💡 Short Explanation

�� Heating provides energy to electrons. �� Electrons gain thermal energy. �� Magnetic fields do not cause thermionic emission.

✅ Correct Answer Explanation

Statement 1 is incorrect because strong magnetic fields do not produce thermionic emission. → Statement 2 is correct because thermionic emission supplies sufficient energy for electron escape. → Statement 3 is correct because the material is suitably heated. → Statement 4 is correct because thermal energy is imparted to free electrons. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 1.
�� Option C → Includes incorrect statement 1.
�� Option D → Includes incorrect statement 1.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each statement using the definition of thermionic emission.

Final Logic:

Only statements 2, 3 and 4 are correct.

🧠 Memory Trick

Thermionic = Heat + Electrons

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

6 In order to enable free electrons to overcome the attractive pull of the metal ions via thermionic emission,

A Each electron must absorb a high-frequency photon

B sufficient thermal energy must be imparted through suitably heating

C The metal must be cooled to absolute zero to reduce vibrations

D an external electric field of 10⁸ V/m must be added to the heat

💡 Short Explanation

�� Thermionic emission is heat-based. �� Electrons gain thermal energy. �� Heating enables escape from the metal surface.

✅ Correct Answer Explanation

In thermionic emission, the metal is heated so that free electrons acquire sufficient thermal energy. → This energy allows them to overcome the attractive force of the positive ions. → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes photoelectric emission.
�� Option C → Cooling reduces electron energy.
�� Option D → Describes field emission rather than thermionic emission.

🎯 Strategy

Used

Elimination

Application:

Identify the option associated with heating.

Final Logic:

Thermionic emission occurs through thermal energy supplied by heating.

🧠 Memory Trick

Heat → Escape

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

7 Incorrect statement about field emission

A It requires applying a very strong electric field.

B The required electric field is on the order of 10⁸ V m⁻¹.

C The process is utilized in devices like spark plugs.

D It works by heating the metal to an extremely high temperature.

💡 Short Explanation

�� Field emission uses strong electric fields. �� Heating is not the mechanism. �� Spark plugs are common applications.

✅ Correct Answer Explanation

Field emission occurs when a very strong electric field extracts electrons from a metal surface. → Typical electric fields are of the order of 10⁸ V m⁻¹. → Heating is associated with thermionic emission, not field emission. → Therefore, Option D is the correct statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of field emission.
�� Option B → Correct order of magnitude.
�� Option C → Correct application.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement describing another emission process.

Final Logic:

Heating belongs to thermionic emission, not field emission.

🧠 Memory Trick

Field = Electric, Not Thermal

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

8 A spark plug utilizes ________ emission, which pulls electrons out of a metal using a strong ________.

A Field, Electric field

B Thermionic, Thermal field

C Photoelectric, Light beam

D Secondary, Magnetic field

💡 Short Explanation

�� Spark plugs use strong electric fields. �� Electrons are extracted from metal surfaces. �� This is field emission.

✅ Correct Answer Explanation

Spark plugs operate by creating strong electric fields. → These fields pull electrons from metallic surfaces. → This process is known as field emission. → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Thermal field is not a standard mechanism.
�� Option C → Spark plugs do not rely on light.
�� Option D → Magnetic fields do not extract electrons in this context.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the device with the emission mechanism.

Final Logic:

Spark plugs use field emission through strong electric fields.

🧠 Memory Trick

Spark = Strong Field

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

9 What mechanism is responsible for an electron gaining sufficient energy to escape the surface attraction during the photoelectric effect?

A Gaining kinetic energy from high-velocity ion collisions.

B Absorbing enough energy from the incident radiation illuminating the surface.

C Accelerating through a strong external electric potential.

D Increasing the internal temperature of the entire metal block uniformly.

💡 Short Explanation

�� Incident radiation transfers energy. �� Electrons absorb photon energy. �� Escape occurs when energy exceeds the work function.

✅ Correct Answer Explanation

In the photoelectric effect, electrons absorb energy from incident electromagnetic radiation. → If the absorbed energy exceeds the work function, electrons escape the metal surface. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ion collisions are not responsible.
�� Option C → External potential does not initiate photoemission.
�� Option D → Heating corresponds to thermionic emission.

🎯 Strategy

Used

Elimination

Application:

Identify the mechanism unique to photoelectric emission.

Final Logic:

Photoelectrons are emitted because they absorb energy from incident radiation.

🧠 Memory Trick

Photon In → Electron Out

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

10 Statements regarding photoelectric terminologies
1. Light falling on a surface always ejects photoelectrons regardless of its frequency.
2. The phenomenon of light emitting electrons is called the photoelectric effect.
3. Emitted charged particles are strictly positively charged.
4. Photo-generated electrons are termed photoelectrons.

A (1), (2)

B (2), (4)

C (1), (3)

D (3), (4)

💡 Short Explanation

�� Frequency must exceed threshold value. �� Emitted particles are electrons. �� Such electrons are called photoelectrons.

✅ Correct Answer Explanation

Statement 1 is incorrect because photoemission occurs only when the incident frequency exceeds the threshold frequency. → Statement 2 is correct because emission of electrons due to incident light is called the photoelectric effect. → Statement 3 is incorrect because the emitted particles are negatively charged electrons. → Statement 4 is correct because emitted electrons are called photoelectrons. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 1.
�� Option C → Includes incorrect statements 1 and 3.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Check each numbered statement using the definition of the photoelectric effect.

Final Logic:

Only statements 2 and 4 are correct.

🧠 Memory Trick

Photo Effect → Photoelectrons

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

11 Incorrect statement about Hertz's 1887 observations

A He investigated the production of electromagnetic waves.

B Light shining on the metal surface facilitated the escape of free charged particles.

C His experiments involved the use of a spark discharge.

D He observed that spark discharges were suppressed by visible light.

💡 Short Explanation

�� Hertz generated electromagnetic waves using spark discharge. �� Ultraviolet light enhanced spark discharge. �� Light facilitated electron emission from metal surfaces.

✅ Correct Answer Explanation

Hertz investigated the production of electromagnetic waves using spark discharges. → He observed that ultraviolet light enhanced spark discharges rather than suppressing them. → The illumination helped charged particles escape from the metal surface. → Therefore, Option D is the correct statement.

❓ Why Other Options Are Incorrect

�� Option A → Correctly describes Hertz's work.
�� Option B → Consistent with the photoelectric effect observation.
�� Option C → Spark discharge was central to the experiment.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement that contradicts Hertz's actual observation.

Final Logic:

Hertz observed enhancement, not suppression, of spark discharge.

🧠 Memory Trick

Hertz + UV = Enhanced Spark

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

12 In Hertz's investigation of electromagnetic waves via spark discharge, the sparks across the detector loop

A were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp

B were completely extinguished by ultraviolet light

C changed color from blue to yellowish-green when hit by visible light

D were unaffected by ultraviolet light but enhanced by infrared light

💡 Short Explanation

�� Ultraviolet light increased spark intensity. �� Arc lamp supplied ultraviolet radiation. �� This observation led to photoelectric studies.

✅ Correct Answer Explanation

Hertz noticed that sparks became stronger when ultraviolet light from an arc lamp illuminated the emitter plate. → This effect indicated that light was helping electrons escape from the metal surface. → Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → UV light enhanced rather than extinguished sparks.
�� Option C → No such color change observation was reported.
�� Option D → UV light produced the effect, not infrared radiation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the historical observation with the correct experimental condition.

Final Logic:

Ultraviolet light from an arc lamp enhanced the spark discharge.

🧠 Memory Trick

Arc Lamp → UV → Stronger Spark

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

13 Hallwachs observed that a negatively charged zinc plate lost its charge under UV light. What does this logically imply about the emitted particles?

A The emitted particles must have a positive charge.

B The emitted particles must be neutral photons.

C The emitted particles must carry a negative charge.

D The plate was absorbing negative particles from the air.

💡 Short Explanation

�� Negative charge decreased from the plate. �� Escaping particles carried away negative charge. �� These particles were electrons.

✅ Correct Answer Explanation

The zinc plate was initially negatively charged. → Under ultraviolet illumination it lost charge. → This can happen only if negatively charged particles leave the surface. → Therefore the emitted particles must be negatively charged electrons. → Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Positive particles leaving would increase negative charge.
�� Option B → Neutral photons cannot explain loss of negative charge.
�� Option D → Absorbing negative particles would increase negative charge.

🎯 Strategy

Used

Elimination

Application:

Analyze charge conservation and the observed discharge.

Final Logic:

Loss of negative charge means emission of negatively charged particles.

🧠 Memory Trick

Negative Charge Lost = Electrons Lost

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

14 Correct statements about Hallwachs' observations on uncharged and positively charged zinc plates
1. An uncharged zinc plate became positively charged when irradiated by UV light.
2. A positively charged zinc plate lost its positive charge under UV light.
3. Positive charge on a positively charged zinc plate was further enhanced under UV light.
4. The zinc plate melted under continuous UV exposure.

A (1), (2)

B (1), (3)

C (2), (4)

D (3), (4)

💡 Short Explanation

�� UV light ejects electrons. �� An uncharged plate becomes positively charged. �� A positively charged plate becomes more positive.

✅ Correct Answer Explanation

Statement 1 is correct because emission of electrons leaves the initially neutral plate positively charged. → Statement 2 is incorrect because a positively charged plate does not lose its positive charge. → Statement 3 is correct because additional electron loss enhances the positive charge. → Statement 4 is incorrect because melting was not observed. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 2.
�� Option C → Includes incorrect statements 2 and 4.
�� Option D → Includes incorrect statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Check each numbered statement against Hallwachs' observations.

Final Logic:

Only statements 1 and 3 are correct.

🧠 Memory Trick

UV Removes Electrons → More Positive

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

15 Lenard's setup featured an ________ tube containing two ________ to carefully measure the photoelectric effect.

A Oxygen-filled, Magnets

B Open-air, Batteries

C Evacuated, Lasers

D Evacuated, Electrodes

💡 Short Explanation

�� Lenard used an evacuated tube. �� Two electrodes were enclosed. �� The setup measured photoelectric current.

✅ Correct Answer Explanation

Lenard investigated photoelectric emission using an evacuated glass tube. → The tube contained two electrodes: an emitter and a collector. → This arrangement enabled accurate measurement of photoelectric current. → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnets were not the primary components.
�� Option B → The experiment was not performed in open air.
�� Option C → Lasers were not used.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Recall the standard Lenard apparatus.

Final Logic:

Lenard's setup consisted of an evacuated tube containing electrodes.

🧠 Memory Trick

Lenard = Vacuum Tube + Electrodes

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

16 If the rate of electrons (charge e) flowing per second from the emitter to the collector is n, the detected current I in the evacuated tube is:

A n + e

B n / e

C n e

D n e²

💡 Short Explanation

�� Current is charge flow per second. �� Each electron carries charge e. �� n electrons per second produce current ne.

✅ Correct Answer Explanation

Current is defined as charge flowing per unit time. → If n electrons pass each second and each carries charge e, I = n × e → Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the formula for current.
�� Option B → Incorrect mathematical relationship.
�� Option D → Contains an unnecessary extra factor of e.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use the definition of electric current.

Final Logic:

Current equals number of charges crossing per second multiplied by charge per particle.

🧠 Memory Trick

Current = Number × Charge

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

17 If the threshold frequency for a metal is 5.0 × 10¹⁴ Hz, what will be the result of irradiating it with light of frequency 4.5 × 10¹⁴ Hz?

A Emits exactly 4.5 × 10¹⁴ electrons

B Zero electron emission

C Emits electrons with 0.5 × 10¹⁴ eV energy

D Saturation current is instantly achieved

💡 Short Explanation

�� Frequency is below threshold frequency. �� No photoelectric emission occurs. �� Intensity cannot compensate for insufficient frequency.

✅ Correct Answer Explanation

Given: Threshold frequency = 5.0 × 10¹⁴ Hz Incident frequency = 4.5 × 10¹⁴ Hz → Since the incident frequency is less than the threshold frequency, electrons cannot acquire enough energy to escape. → Therefore no photoelectric emission takes place. → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Number of emitted electrons is zero.
�� Option C → No electrons are emitted.
�� Option D → Saturation current cannot occur without photoemission.

🎯 Strategy

Used

Substitution

Application:

Compare the incident frequency directly with threshold frequency.

Final Logic:

f < f₀ ⇒ No photoelectric emission.

🧠 Memory Trick

Below Threshold = No Emission

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

18 Match List I (Concept) with List II (Description) regarding threshold frequency.

List I	List II (Correct Match)
(1) Incident light frequency < Threshold frequency	(a) Determines the exact threshold frequency limit
(2) Threshold frequency	(b) No electrons are emitted at all
(3) Nature of emitter material	(c) Minimum value for emission to occur
(4) Incident light frequency > Threshold frequency	(d) Emission of electrons takes place

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

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

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

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

💡 Short Explanation

�� Below threshold frequency no emission occurs. �� Threshold frequency is the minimum required frequency. �� Material nature determines threshold frequency.

✅ Correct Answer Explanation

The correct matching is: → (1) Incident light frequency < Threshold frequency → (b) No electrons are emitted at all → (2) Threshold frequency → (c) Minimum value for emission to occur → (3) Nature of emitter material → (a) Determines the exact threshold frequency limit → (4) Incident light frequency > Threshold frequency → (d) Emission of electrons takes place → These statements summarize the threshold frequency condition for photoelectric emission.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly reverses emission conditions.
�� Option C → Incorrectly defines threshold frequency.
�� Option D → Multiple mismatches occur.

🎯 Strategy

Used

Option Grouping

Application:

Match each threshold-frequency concept with its proper description.

Final Logic:

Only Option A provides all correct pairings.

🧠 Memory Trick

Below → No Emission, Above → Emission

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

19 Unlike heavy metals such as zinc and cadmium, alkali metals like caesium and rubidium

A require high-energy X-rays to emit photoelectrons

B are sensitive even to visible light to cause electron emission

C only emit photoelectrons in the presence of an electric field

D block the emission of negatively charged particles

💡 Short Explanation

�� Alkali metals have low work functions. �� Visible light can cause photoemission. �� They are highly photosensitive.

✅ Correct Answer Explanation

Caesium and rubidium possess relatively low work functions. → Therefore visible light provides sufficient energy for electron emission. → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → X-rays are unnecessary.
�� Option C → Electric fields are not required to initiate photoemission.
�� Option D → They emit electrons rather than blocking them.

🎯 Strategy

Used

Elimination

Application:

Compare the photoelectric properties of alkali metals and heavy metals.

Final Logic:

Low work functions make alkali metals sensitive to visible light.

🧠 Memory Trick

Caesium & Rubidium = Visible Light Metals

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

20 Why do zinc, cadmium, and magnesium respond only to ultraviolet light for electron emission, while sodium and potassium can use visible light?

A Visible light has a shorter wavelength than ultraviolet light.

B Ultraviolet light is inherently negatively charged, attracting the positive ions.

C Alkali metals have much higher work functions than zinc or cadmium.

D Ultraviolet light has short wavelength, thus higher energy is necessary to overcome the higher barriers of the former metals.

💡 Short Explanation

�� Ultraviolet light has higher photon energy. �� Zinc, cadmium and magnesium have higher work functions. �� Alkali metals have lower work functions.

✅ Correct Answer Explanation

Photon energy is given by: E = hν = hc/λ → Ultraviolet light has shorter wavelength and therefore higher energy than visible light. → Zinc, cadmium and magnesium require this higher energy to overcome their larger work functions. → Sodium and potassium possess lower work functions, so visible light can eject electrons from them. → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Visible light has a longer wavelength than ultraviolet light.
�� Option B → Ultraviolet radiation is electrically neutral.
�� Option C → Alkali metals actually have lower work functions.

🎯 Strategy

Used

Elimination

Application:

Compare photon energy and work function concepts.

Final Logic:

Higher work function metals require the higher-energy ultraviolet photons.

🧠 Memory Trick

Higher Work Function → Need UV

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

CUET UG Physics Booster Test 2-Electron Emission and Photoelectric Observations

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