Amagnesium surface has a work function of 3.60 ev. electromagnetic waves with a wavelength of 320 nm strike the surface and eject electrons. find the maximum kinetic energy of the ejected electrons. express your answer in electron volts.

K(maximum) = 0.284 eV

Explanation:

Given data

work function ω = 3.60 eV = 3.60 × 1.6 × = 5.76 × J

wavelength λ = 320 nm  = 320 × m

to find out

maximum kinetic energy

solution

we know that speed of light is =  3 × m/s

and planck constant h = 6.63 × J.s

we will apply here Einstein's photo elecric equation that is

hc/λ  =  ω + K(maximum)

put here all these value we get

6.63 × ×  3 ×  / 320 × =   5.76 ×  + K(maximum)

so

K(maximum) = 6.63 × ×  3 ×  / 320 × - 5.76 ×

K(maximum) = 0.4556 ×  J

K(maximum) = 0.4556 ×  / 1.6  ×

K(maximum) = 0.284 eV

b

explanation: