MyCompetitiveExams Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
Photoemission occurs ONLY if photon energy E ≥ work function Φ₀! 2.20 eV is greater only than 2.14 eV (Cs), so Cs only in 5 seconds!
According to Einstein's quantum theory of the photoelectric effect, an incident photon transfers its entire energy E = hν to a single conduction electron. Photoelectric emission can occur only if this energy exceeds the minimum energy required to liberate the electron from the metal surface (the work function Φ₀):
Compare the incident radiation energy E = 2.20 eV with the work functions of all three photosensitive alkali metals:
Only Caesium satisfies the condition E ≥ Φ₀, meaning photoelectrons are emitted from the Caesium surface only.
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Sodium has the highest work function (2.75 eV), requiring photons of at least 2.75 eV.
Accurate deduction: only Caesium possesses a work function (2.14 eV) lower than the incident energy (2.20 eV).
Both metals have work functions exceeding 2.20 eV.
Potassium requires 2.30 eV; incident photons have only 2.20 eV.
A student who mistakenly believes "the metal with the largest work function emits photoelectrons" will pick Na. In reality, lower work function means electrons are held less tightly and escape more easily.
"Photoelectric emission is an all-or-nothing instantaneous threshold phenomenon occurring strictly when E ≥ Φ₀."
Test your speed and accuracy in the authentic 200-minute timed NTA simulator.
Spotted a mistake in this solution? Report an error ([email protected])
Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.