Respuesta :
Answer:
given,
light wavelength = 209 nm
maximum kinetic energy = 3.57 × 10⁻¹⁹ J
using photo electric equation
[tex]\dfrac{hc}{\lambda} = \varphi+ K_{max}[/tex]
[tex]\varphi =\dfarc{hc}{\lambda}- K_{max}[/tex]
= [tex]\dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{209 \times 10^{-9}} - 3.57\times 10^{-19}[/tex]
=5.94× 10⁻¹⁹ J
wavelength of light that should be used for double maximum K.E
[tex]\dfrac{hc}{\lambda'} = \varphi+ 2K_{max}[/tex]
[tex]\lambda' =\dfrac{hc}{ \varphi+ 2K_{max}}[/tex]
[tex]\lambda' =\dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{ 5.94\times 10^{-19}+ 2\times 3.57 \times 10^{-19}}[/tex]
= 151.94 × 10⁻⁹ m = 151.94 m
