Homework Answers
(a)
Φ = 2.0 eV
We know,
(h*c)/λ = 2.0 eV
(4.135 * 10-15 * 3.0 * 10^8) / λ = 2.0
λ = 6.2025 * 10^-7 m
(b)
Energy of photoelectron = (h*c)/λ = (4.135 * 10^-15 *
3.0 * 10^8) / (200 * 10^-9) = 6.2025 eV
Total Energy = 6.2025 * 1.6 * 10^-19 J
No of Photoelectron = 40 / ( 6.2025 * 1.6 * 10^-19)
No of Photoelectron = 4.03 * 10^19
(c)
Ekmax = (h*c)/λ - Φ
Ekmax = (4.135 * 10-15 * 3.0 * 10^8) / (200 * 10^-9) - 2.0
Ekmax = 4.2025 eV
For an individual photoelectron,
Ekmax = (4.2025 * 1.6 * 10^-19)/( 4.03 * 10^19) J
Ekmax = 1.67 * 10^-38 J
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