If the work function of a metal is $3.7 e V$ . When a photon of energy $5.2 e V$ falls on the metal surface then the kinetic energy of ejected photoelectrons will be ______.

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Solution

Step 1. Given data:
Work function = $3.7 e V$
The energy of photon = $5.2 e V$
Step 2. Formula used:
The kinetic energy of ejected electrons $(K . E)$ = Energy of photon $(E_{P H})$ - work function $(Φ)$
$K . E = E_{P H} - Φ$
Step 3. Calculating kinetic energy
Putting the given values, we get
$K . E = 5.2 - 3.7 = 1.5 e V$
Thus, the kinetic energy of ejected photoelectrons will be 1.5 eV.

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If the work function of a metal is 3.7eV. When a photon of energy 5.2eV falls on the metal surface then the kinetic energy of ejected photoelectrons will be ______.

If the work function of a metal is $3.7 e V$ . When a photon of energy $5.2 e V$ falls on the metal surface then the kinetic energy of ejected photoelectrons will be ______.