Question

The threshold frequency for a metal is 6×10¹⁴ s⁻¹ . The kinetic energy of the electron emitted when the radiation of frequency 2×10¹⁵ s⁻¹ hits the metal is

a) 9.3×10⁻²⁰J
b) -6×10¹⁴ J
c) 92.8×10⁻³³ J
d) 9.3×10⁻¹⁹ J

Answer 1

Answer:

threshold frequency = 6×10¹⁴s⁻¹

radiated frequency =  2×10¹⁵ s⁻¹

kinetic energy = ?

h (radiated frequency - threshold frequency) = kinetic energy

h = planks constant = 6.626 x 10^-34 Js

so substituting all the numerical values we get

6.626 x 10^-34 x (2×10¹⁵ -  6×10¹⁴) = kinetic energy

K.E = 6.626 X 10^-34 (20 x 10¹⁴- 6 x 10¹⁴)

= 6.626 x 10^-34 x 10^14 ( 20 - 6)

= 6.626 x ( 14) x 10^(-34 + 14)

= 92.794 x 10^-20 J

which is approximatly 9.3×10⁻²⁰J

thus K.E is option a) 9.3×10⁻²⁰J

Explanation:

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