When light with a wavelength of 230 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.35 × 10 − 19 3.35 × 10 − 19 J. Determine the wavelength of light that should be used to triple the maximum kinetic energy of the electrons ejected from this surface.

Question

contestada

Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2019-08-27T07:36:42+02:00

Answer:

wavelength is 121.58 nm

Explanation:

given data

wavelength = 230 nm = 230 x [tex]10^{-9}[/tex] m

kinetic energy = 3.35 × [tex]10^{-19}[/tex] J

to find out

wavelength of light that triple the maximum kinetic energy of the electron

solution

we know here Einstein's photo electric equation that is given below

hc / λ = φ + K(maximum)

and here φ will be

φ = (hc) / λ - K(max imum)

put here h = 6.626 x [tex]10^{-34}[/tex] J.s

and 3.0 x [tex]10^{8}[/tex] m/s

so φ = (6.626 x [tex]10^{-34}[/tex] )( 3.0 x [tex]10^{8}[/tex] ) / ( 230 x [tex]10^{-9}[/tex] ) - 3.35 x[tex]10^{-19}[/tex]

φ = 6.2995 x [tex]10^{-19}[/tex] J

so here we have given triple the maximum kinetic energy

so

hc / λ = φ + K(maximum)

λ = hc / ( φ + K(maximum) )

λ = (6.626 x [tex]10^{-34}[/tex] )( 3.0 x [tex]10^{8}[/tex] ) / ( 6.2995 x [tex]10^{-19}[/tex] + 3 ( 3.35 x[tex]10^{-19}[/tex] ))

λ = 1.215817 × [tex]10^{-7}[/tex]

so wavelength is 121.58 nm