The energy of emitted photoelectrons from a metal is 0.9 eV. The work function of the metal is 2.2 eV. Then the energy of the incident photon is :

0.9 eV

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2. 2 eV

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4. 4 eV

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3.1 eV

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Solution

The correct option is D 3.1 eV
Energy of incident photon $=$ energy of emitted photoelectrons $+$ work function
$= 2.9 e V + 2.2 e V$
$= 3.1 e V$
So, the answer is (D).

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