Light of wavelength 2000 $\circ A$ falls on an aluminium surface. Work function of the aluminium is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons?

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Solution

Given:
Incident wavelenth, $λ_{i n c} = 2000 0 A$

Hence;
Energy ( $E_{i n c}$ ) of incident light can be given as:
$E_{i n c} = \frac{12400 e V 0 A}{λ (0 A)}$

$E_{i n c} = \frac{12400}{2000}$

$= 6.2 e V$

whereas work function of Aluminium;

$Φ_{A L} = 4.2 e V$

Hence; Using Photo-electric effective

$E_{i n c} = Φ + (K E)_{m a x}$

Hence, $(K . E)_{m a x} = E_{i n c} - Φ_{A L}$

$= 6.2 - 4.2$

$= 2 e V$

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Light of wavelength 2000 ∘A falls on an aluminium surface. Work function of the aluminium is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons?

Given: Incident wavelenth, λinc=20000AHence;Energy (Einc) of incident light can be given as: Einc=12400eV 0Aλ(0A)Einc=124002000 =6.2 eVwhereas work function of Aluminium;ΦAL=4.2 eVHence; Using Photo-electric effectiveEinc=Φ+(KE)maxHence, (K.E)max=Einc−ΦAL =6.2−4.2 =2 eV

Light of wavelength 2000 $\circ A$ falls on an aluminium surface. Work function of the aluminium is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons?