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

Answer:

given,

light wavelength = 209 nm

maximum kinetic energy = 3.57 × 10⁻¹⁹ J

using photo electric equation

$\dfrac{hc}{\lambda} = \varphi+ K_{max}$

$\varphi =\dfarc{hc}{\lambda}- K_{max}$

           = $\dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{209 \times 10^{-9}} - 3.57\times 10^{-19}$

         =5.94× 10⁻¹⁹ J

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

$\dfrac{hc}{\lambda'} = \varphi+ 2K_{max}$

$\lambda' =\dfrac{hc}{ \varphi+ 2K_{max}}$

$\lambda' =\dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{ 5.94\times 10^{-19}+ 2\times 3.57 \times 10^{-19}}$

              = 151.94 × 10⁻⁹ m = 151.94 m