Question

Calculate the kinetic energy of the ejected electron then ultraviolet radiation of frequency 1.6 x 10¹⁵ s
-¹


strikes the surface of potassium metal. Threshold frequency of Potassium is 5 x 10¹⁴ s
-¹ (h = 6.63 x 10-³⁴ J s). ​

Answer 1

Answer:

K. E =h (v - v')

h=planks constant

v=frequency

v'=threshold frequency

=>k. E=6.63 × 10-³⁴([1.6-0.5] × 10^15

=>K. E=7.29 × 10^-19 joules.....

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Answer 2

Answer:

K.E. = hv-hvo

where, hvo= threshold frequence

hv = frequency of light

K.E. = 6.626x10-34 Js(1.6x1015 s-1- 5x1014 s-1)

         = 7.2886 x10-18 J