Question

Photoelectrons are removed with kinetic energy 1.864×10^-21 J, when photons of light with energy 4.23×10^-19 J fall on the metal. What is the minimum energy required per mole to remove an electron from potassium metal?

Answer 1

Energy of one photon is given by the expression E = hV
where, h = PlancKs constant = 6.636 x 10-34J s-1
V= Frequency of radiation = 5x 1014 Hz= 5 x 1014 s-1.
E= (6.636 x 1034 J s-1 x 5 x 1014 s-1)= 33.18 x 1020J
One mole of photon = 6.023 x 1023photons
Therefore, Energy of one mole of photon = (6.023 x 1023 x 33.18 x 10-20 J) = 199.84 x 103 J
Energy of one mole of photon = 199.84 kJ 

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Answer 2

Answer:

Explanation:

Energy of one photon is given by the expression E = hV

where, h = PlancKs constant = 6.636 x 10^-34J s-1

V= Frequency of radiation = 5x 10^14 Hz= 5 x 10^14 s-1.

E= (6.636 x 10^34 J s-1 x 5 x 10^14 s-1)= 33.18 x 10^20J

One mole of photon = 6.023 x 10^23photons

Therefore, Energy of one mole of photon = (6.023 x 10^23 x 33.18 x 10^-20 J) = 199.84 x 10^3 J

Energy of one mole of photon = 199.84 kJ