A common confusion here: reading Hertz's and Lenard's experiments as measurements of electron energy. They were not. Neither wrote an energy equation. They established that light of high enough frequency liberates negative charge from a metal, and which surface it leaves.
Hertz, 1887. While studying the spark discharge that produced his electromagnetic waves, Hertz noticed the discharge across the gap occurred more readily — at lower applied voltage — when ultraviolet light fell on the negative electrode. UV was helping electrons escape the metal, so a smaller field sufficed. The finding was incidental: he was building a radio-wave experiment.
Hallwachs, 1888. A clean zinc plate joined to a gold-leaf electroscope, lit with UV. A negatively charged plate lost its charge; a neutral plate acquired positive charge; an already-positive plate became more positive. All three point one way: the plate loses negative particles.
Lenard, 1900. Two electrodes in an evacuated tube with a potential difference across them. Shine UV on the emitter plate and current flows in the external circuit; stop the light and the current stops at once. Charged particles crossing the vacuum complete the circuit — emitted by the illuminated plate, gathered by the other. Illuminating the collector instead does nothing.
The particles were identified as electrons when their charge-to-mass ratio matched Thomson's cathode-ray value, and the effect was named photoelectric. Hallwachs and Lenard also recorded a frequency selectivity: alkali metals — lithium, sodium, potassium, caesium — respond to visible light, while zinc, cadmium and magnesium need ultraviolet (NCERT Class 12 Physics, Chapter 11, page 275).
In NEET, this topic is tested as whose observation was what and which plate was illuminated. The energy bookkeeping — work function, threshold, stopping potential — belongs to the Einstein-equation topics.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Hertz first noticed the effect later known as the photoelectric effect while he was investigating:
Answer: C. Hertz's 1887 work was on electromagnetic waves; the effect turned up incidentally, as a change in his spark-gap behaviour under ultraviolet light (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because the charge-to-mass ratio measurement is Thomson's cathode-ray work; it later identified the emitted particles but was not the experiment in which the emission was first noticed.
Why B is wrong: B is wrong because the zinc-plate-and-electroscope experiment is Hallwachs' (1888), a year after Hertz.
Why D is wrong: D is wrong because the evacuated two-electrode tube is Lenard's apparatus (1900), not Hertz's spark gap.
Ultraviolet light is shone on the negative electrode of a spark gap across which a slowly rising voltage is applied. Compared with the same gap in darkness, the discharge now occurs:
Answer: A. The ultraviolet light helps electrons leave the negative electrode, so fewer volts are needed to start the discharge — this is precisely Hertz's observation (NCERT Class 12 Physics, Chapter 11, page 275).
Why B is wrong: B is wrong because it reverses the observation: the ultraviolet light assists the escape of electrons rather than hindering it, so the voltage needed falls.
Why C is wrong: C is wrong because although light itself is uncharged, it supplies the energy an electron needs to leave the surface, so the discharge condition does change.
Why D is wrong: D is wrong because the assistance is present only while the light falls on the electrode; removing the light removes the effect.
The observation that a clean zinc plate connected to a gold-leaf electroscope changes its charge state under ultraviolet illumination is credited to:
Answer: D. Hallwachs carried out the zinc-plate electroscope experiments in 1888, following up Hertz's spark-gap observation (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because Hertz's 1887 observation concerned the spark discharge itself; no electroscope was involved.
Why B is wrong: B is wrong because Lenard's contribution was the evacuated two-electrode tube showing a light-controlled current, around 1900.
Why C is wrong: C is wrong because Einstein's 1905 contribution was the theoretical explanation, not an observation on charged plates.
An uncharged, clean zinc plate attached to an electroscope is illuminated with ultraviolet light. The plate:
Answer: B. The ultraviolet light ejects negative particles, so what remains on the plate is a net positive charge — Hallwachs' result for an initially neutral plate (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because emission does not require the plate to be pre-charged; it loses negative particles from the neutral state and therefore ends up positive.
Why C is wrong: C is wrong because the sign is backwards: negative particles leave the plate, so the plate cannot become more negative.
Why D is wrong: D is wrong because zinc responds to ultraviolet and not to visible light, so the stated condition is inverted.
Lenard's tube holding the emitter and collector plates was evacuated. The main reason is that in air:
Answer: D. The particles freed at the emitter have to cross to the collector for a light-controlled current to appear, and gas molecules in the path scatter them — which is why the experiment specifies an evacuated glass tube (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because ultraviolet light does propagate through air; partial absorption of the beam is not the reason for evacuating the tube.
Why B is wrong: B is wrong because Lenard's apparatus read a current in an external circuit, not an electroscope deflection; the electroscope belongs to Hallwachs' arrangement.
Why C is wrong: C is wrong because the external source maintains the potential difference whatever gas lies between the electrodes; evacuation is about the particles' free passage, not about holding the voltage.
In a Lenard-type tube the ultraviolet beam is moved so that it falls on the collector plate rather than the emitter plate, with everything else unchanged. The current in the external circuit:
Answer: A. Emission happens at the illuminated surface, and only particles leaving the emitter are driven across by the applied field; illuminating the collector produces no such flow (NCERT Class 12 Physics, Chapter 11, page 275).
Why B is wrong: B is wrong because what matters is which surface the light strikes, not that both are metal; the field carries particles from emitter to collector, not the other way.
Why C is wrong: C is wrong because the emitter is no longer illuminated at all — the beam was moved, not duplicated — so there is no second contribution to add.
Why D is wrong: D is wrong because particles freed at the collector are drawn straight back to it by its own positive potential, so no steady current flows in either direction.
In the observations of Hallwachs and Lenard on the response of different metals to light:
Answer: B. The alkali metals respond to visible light, whereas zinc, cadmium and magnesium respond only to ultraviolet (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because the two groups are swapped: it is the alkali metals that are sensitive to visible light.
Why C is wrong: C is wrong because brightness does not substitute for the required colour of the light — zinc stays inert under intense visible illumination.
Why D is wrong: D is wrong because the alkali metals do respond to visible light, which is exactly why they were singled out in these observations.
What established that the particles emitted from an illuminated metal surface in these experiments were electrons?
Answer: C. The emission experiments showed only that negative particles left the surface; matching their charge-to-mass ratio with Thomson's cathode-ray value identified them as electrons (NCERT Class 12 Physics, Chapter 11, page 275).
Why A is wrong: A is wrong because a change in spark-gap voltage shows that escape from the metal was made easier; it says nothing about the identity of whatever escaped.
Why B is wrong: B is wrong because prompt stopping shows the emission is driven by the light rather than by stored charge, but it does not identify the particle.
Why D is wrong: D is wrong because Einstein's equation came in 1905, after the identification, and describes the energy of the emitted electrons rather than establishing what they are.
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Given.
A clean zinc plate is given a small negative charge and connected to a gold-leaf electroscope; the leaf stands out. Ultraviolet light is then shone steadily on the plate. The leaf first collapses to zero, and under continued illumination it diverges again.
Required.
Account for both stages, and state the sign of the plate's charge at the end.
Concept.
Ultraviolet light on zinc ejects negative particles from the surface (Hallwachs' observation). Separately: a gold-leaf electroscope responds to the magnitude of the charge on the plate, not to its sign — it cannot tell positive from negative.
Formula.
None. This topic has no formula of its own in the corpus; the item is solved by tracking the plate's charge budget.
Substitution.
Take the initial charge as −q, with q > 0. Let the total negative charge carried away by emission up to any instant be n.
Calculation.
Net charge on the plate is (−q + n). While n < q the plate is still negative and its magnitude |−q + n| is shrinking, so the leaf falls. At n = q the net charge is zero and the leaf collapses fully. For n > q the net charge is positive and growing, so the magnitude rises again and the leaf re-diverges. No physical constants enter this item, so no question of exactness or significant figures arises.
Final answer.
Stage 1 is the neutralisation of the original negative charge by the loss of negative particles; stage 2 is the build-up of positive charge on the plate once the original charge has gone. The final charge is positive.
Common trap.
Reading the second divergence as a return to negative charge. The electroscope is sign-blind, so divergence alone never tells you the sign — you have to follow which way charge is moving. A second version of the same error is imagining the ultraviolet light delivers positive charge; nothing positive arrives, negative charge departs.
Similar NEET-style question.
A clean zinc plate carrying a small positive charge is attached to an electroscope and illuminated with ultraviolet light. The leaf divergence: (a) collapses, then rises; (b) increases steadily; (c) stays constant; (d) collapses to zero and stays there. Answer: (b) — the plate is already positive, and continued loss of negative particles makes it more positive, in line with the NCERT statement that a positively charged plate becomes more positive under ultraviolet light.
Emission of electrons from a metal when light of suitable frequency falls on it. Discovered by Hertz, studied by Lenard. Below threshold frequency ν₀, no electrons emitted regardless of intensity.
-- NCERT Class 12 Physics, Ch. 11, p. 276More in Dual Nature of Matter and Radiation: 2 exam traps and mistakes · 3 formulas · 2 question patterns from its other lessons.
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