Question

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Answer 1

The liberated electron will have 10.28 eV kinetic energy.

Given that light of frequency 3.62 × $10^{15}$ Hz strikes a surface of copper metal with work function($W_{0}$  = 4.70 eV).

The incident (photon) energy given as hf (Planck's Constant time's frequency) minus the energy that "binds" the electron to the metal (the work function) will equal the maximum kinetic energy of the emitted electron.

K.E. = hf − ϕ

Where K = Kinetic energy

h = Planck's constant = 6.62607015 × $10^{-34}$ J s

f = frequency = 3.62 × $10^{15}$ Hz

ϕ = 4.70 eV = 4.7 × 1.6 × $10^{-19}$ = 7.52 × $10^{-19}$

K.E. = 6.62607015 × $10^{-34}$ × 3.62 × $10^{15}$ - 7.52 × $10^{-19}$

       = 6.62607015 × 3.62 × $10^{-19}$ - 7.52 × $10^{-19}$

       = 23.98 × $10^{-19}$ - 7.52 × $10^{-19}$

K.E. = 16.46 × $10^{-19}$ J = 10.28 eV

To know more about work function refer to: brainly.com/question/14287500

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