A magnesium 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.

Respuesta :

Answer:

K(maximum) = 0.284 eV

Explanation:

Given data

work function ω = 3.60 eV = 3.60 × 1.6 ×[tex]10^{-19}[/tex] = 5.76 ×[tex]10^{-19}[/tex] J

wavelength λ = 320 nm  = 320 ×[tex]10^{-9}[/tex] m

to find out

maximum kinetic energy

solution

we know that speed of light is =  3 ×[tex]10^{8}[/tex] m/s

and planck constant h = 6.63 ×[tex]10^{-34}[/tex] 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 ×[tex]10^{-34}[/tex] ×  3 ×[tex]10^{8}[/tex]  / 320 ×[tex]10^{-9}[/tex] =   5.76 ×[tex]10^{-19}[/tex]  + K(maximum)

so

K(maximum) = 6.63 ×[tex]10^{-34}[/tex] ×  3 ×[tex]10^{8}[/tex]  / 320 ×[tex]10^{-9}[/tex] - 5.76 ×[tex]10^{-19}[/tex]

K(maximum) = 0.4556 ×[tex]10^{-19}[/tex]  J

K(maximum) = 0.4556 ×[tex]10^{-19}[/tex]  / 1.6  ×[tex]10^{-19}[/tex]

K(maximum) = 0.284 eV