Question:medium

When a photon of energy hν falls on a photosensitive metallic surface (work function hν0), electrons are emitted from the metallic surface. It is possible to say that:

Updated On: Mar 27, 2026
  • All have the same K.E.=hν−hν$_0​$
  • Maximum K.E.=hν−hν$_0$​
  • Maximum K.E.=hν
  • K.E.=hν$_0$​
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The Correct Option is B

Solution and Explanation

Electron emission occurs from a photosensitive metal upon photon incidence if the photon energy \( hu \) exceeds the metal's work function \( hu_0 \). The photon's energy is partitioned into overcoming the work function and providing kinetic energy to the emitted electron.

The energy balance is expressed as:

\[ hu = hu_0 + KE_{\text{max}} \]

where \( KE_{\text{max}} \) denotes the maximum kinetic energy of the liberated electrons.

Rearranging yields:

\[ KE_{\text{max}} = hu - hu_0 \]

This equation indicates that the highest kinetic energy of emitted electrons equals the incident photon energy minus the material's work function.

Consequently, the accurate statement among the given choices is: Maximum K.E.=hu−hu_0​.

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