In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of 1.2 V. Find energy of incident photon.

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Solution

Given,
Work function, $W_{o} = 3.5 e V$

Potential, $V = E_{k} = 1.2 e V$

According to Einstein’s photoelectric equation,

$h ν = W_{o} + \frac{1}{2} m v^{2}$
$h ν = 3.5 + 1.2$

$h ν = 4.7 e V$

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In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of 1.2 V. Find energy of incident photon.

Given,Work function, Wo=3.5eV Potential, V=Ek=1.2eV According to Einstein’s photoelectric equation, hν=Wo+12mv2hν=3.5+1.2 hν=4.7eV

Given,
Work function, $W_{o} = 3.5 e V$

Potential, $V = E_{k} = 1.2 e V$

According to Einstein’s photoelectric equation,

$h ν = W_{o} + \frac{1}{2} m v^{2}$
$h ν = 3.5 + 1.2$

$h ν = 4.7 e V$