Question

When light with a wavelength of 209 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.57 × 10^-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer 1

given,

light wavelength = 209 nm

maximum kinetic energy = 3.57 × 10⁻¹⁹ J

using photo electric equation

`(hc)/(\lambda) = \varphi+ K_(max)`

`\varphi =\dfarc{hc}{\lambda}- K_(max)`

=
`(6.626* 10^(-34)* 3 * 10^8)/(209 * 10^(-9))  - 3.57* 10^(-19)`

=5.94× 10⁻¹⁹ J

wavelength of light that should be used for double maximum K.E

`(hc)/(\lambda') = \varphi+ 2K_(max)`

`\lambda' =(hc)/( \varphi+ 2K_(max))`

`\lambda' =(6.626* 10^(-34)* 3 * 10^8)/( 5.94* 10^(-19)+ 2* 3.57 * 10^(-19))`

= 151.94 × 10⁻⁹ m = 151.94 m