IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

edison

Problem) Calculate the kinetic energy of the electron when ultra violet radiation of frequency 1.6 * 10¹⁵/s

strikes the surface of potassium metal .Threshold

frequency of potassium is 5 * 10¹⁴ /s

which principle is applicable and why?

Solution) It is a problem of photoelectric effect (PEE)

because of the phenomenon involved

Kinetic energy of photo electron is where given by

K.E = hf - hf₀

Where, f = frequency of the incident photon

and f₀ = Threshhold frequency for PEE to take place.

h = planck's constant.

Thus, KE = h x (1.6 * 10¹⁵ - 5 * 10¹⁴ )

or, KE = h x 9 x 10¹⁴

or, KE = 6.626068 × 10^-34 x 9 x 10¹⁴ J

or, KE = 6.626068 × 10^-34 x 9 x 10¹⁴ / (1.6 x 10^-19) eV

solve to get KE in eV.