Question

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.

Answer 1

K(maximum) = 0.284 eV

Step-by-step explanation:

Given data

work function ω = 3.60 eV = 3.60 × 1.6 ×
`10^(-19)` = 5.76 ×
`10^(-19)` J

wavelength λ = 320 nm = 320 ×
`10^(-9)` m

to find out

maximum kinetic energy

solution

we know that speed of light is = 3 ×
`10^(8)` m/s

and planck constant h = 6.63 ×
`10^(-34)` 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 ×
`10^(-34)` × 3 ×
`10^(8)` / 320 ×
`10^(-9)` = 5.76 ×
`10^(-19)` + K(maximum)

so

K(maximum) = 6.63 ×
`10^(-34)` × 3 ×
`10^(8)` / 320 ×
`10^(-9)` - 5.76 ×
`10^(-19)`

K(maximum) = 0.4556 ×
`10^(-19)` J

K(maximum) = 0.4556 ×
`10^(-19)` / 1.6 ×
`10^(-19)`

K(maximum) = 0.284 eV