Question

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.

Answer 1

wavelength is 121.58 nm

Step-by-step explanation:

given data

wavelength = 230 nm = 230 x
`10^(-9)` m

kinetic energy = 3.35 ×
`10^(-19)` 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
`10^(-34)` J.s

and 3.0 x
`10^(8)` m/s

so φ = (6.626 x
`10^(-34)` )( 3.0 x
`10^(8)` ) / ( 230 x
`10^(-9)` ) - 3.35 x
`10^(-19)`

φ = 6.2995 x
`10^(-19)` 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
`10^(-34)` )( 3.0 x
`10^(8)` ) / ( 6.2995 x
`10^(-19)` + 3 ( 3.35 x
`10^(-19)` ))

λ = 1.215817 ×
`10^(-7)`

so wavelength is 121.58 nm