Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

Question

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Answer 1

Work function of plate $ϕ = 1.9 e V$
Wavelength of incident photon $λ = 5000 A^{o} = 500 n m$
Energy of the incident photon $E_{i n c i d e n t} = \frac{1 2 4 0}{λ ( i n n m )} e V$
$⟹ E_{i n c i d e n t} = \frac{1 2 4 0}{5 0 0} = 2.48 e V$
Maximum kinetic energy of photon emitted $K . E_{m a x} = E_{i n c i d e n t} - ϕ$
$⟹ K . E_{m a x} = 2.48 - 1.9 = 0.58 e V$

Answer 2

Work function of plate

ϕ = 1.9 e V

Wavelength of incident photon

λ = 5000 A^{o} = 500 n m

Energy of the incident photon

E_{i n c i d e n t} = \frac{1 2 4 0}{λ ( i n n m )} e V

⟹ E_{i n c i d e n t} = \frac{1 2 4 0}{5 0 0} = 2.48 e V

Maximum kinetic energy of photon emitted

K . E_{m a x} = E_{i n c i d e n t} - ϕ

⟹ K . E_{m a x} = 2.48 - 1.9 = 0.58 e V