Question

. Barium has a work function of 2.48eV. What is the maximum kinetic energy if the metal is illuminated by UV light of wavelength 365nm?​

Answer 1

The maximum kinetic energy is 0.91 eV

What is Work Function of a Metal ?

A Work Function of a Metal is the minimum energy required to liberate an electron from a metal surface. From the formula;

hF = hf + E

• hF = energy for liberation or illumination

• hf = work function = 2.8 eV

• E = kinetic energy

Given that a Barium has a work function of 2.48eV. What is the maximum kinetic energy ?

Convert eV to Joule by multiplying by 1.6 x $10^{-19}$

hf = 2.48 x 1.6 x $10^{-19}$

hf = 3.968 x $10^{-19}$ J

If the metal is illuminated by UV light of wavelength 365nm,

E = hF = hc/λ

Where

• C = speed of light = 3 x $10^{8}$ m/s

• h = Planck constant = 6.6 x $10^{-34}$ Js

• λ = wavelength = 365 x $10^{-9}$ m

Substitute all the parameters

hF = (6.6 x $10^{-34}$ x  3 x $10^{8}$) / 365 x $10^{-9}$

hF = 1.98 x $10^{-25}$ /  365 x $10^{-9}$

hF = 5.42 x $10^{-19}$ Joule

The maximum kinetic energy will be

hF = hf + E

Make E the subject of formula

E = hF - hf

E = 5.42 x $10^{-19}$ - 3.968 x $10^{-19}$

E = 1.452 x $10^{-19}$ J

Convert Joule back to eV by dividing by 1.6 x $10^{-19}$

E = 1.452 x $10^{-19}$ / 1.6 x $10^{-19}$

E = 0.9075 eV

Therefore, the maximum kinetic energy is 0.91 eV approximately

Learn more about Work Function here: brainly.com/question/14144692

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