Question

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 4.17×1014 s−1. E0= J With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of 245 nm? KEelectron= J

Answer 1

Answer:

Explanation:

Threshold frequency = 4.17 x 10¹⁴ Hz .

minimum energy required = hν where h is plank's constant and ν is frequency .

E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴

= 27.52 x 10⁻²⁰ J .

wavelength of radiation falling = 245 x 10⁻⁹ m

Energy of this radiation = hc / λ

c is velocity of light and  λ  is wavelength of radiation .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹

= .08081 x 10⁻¹⁷ J

= 80.81 x 10⁻²⁰ J

kinetic energy of electrons ejected = energy of falling radiation - threshold energy

= 80.81 x 10⁻²⁰  - 27.52 x 10⁻²⁰

= 53.29 x 10⁻²⁰ J .