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.

wavelength is 121.58 nm

Explanation:

given data

wavelength = 230 nm = 230  x m

kinetic energy = 3.35 ×   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 J.s

and 3.0 x m/s

so φ = (6.626 x )( 3.0 x   ) / ( 230 x )  -  3.35 x

φ = 6.2995  x  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 )( 3.0 x   )  / ( 6.2995  x  + 3 ( 3.35 x ))

λ  = 1.215817 ×

so wavelength is 121.58 nm

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation: