Respuesta :
Answer:
[tex]K.E.=9.4657\times 10^{-20}\ J[/tex]
Explanation:
Using the expression for the photoelectric effect as:
[tex]E=h\nu_0+\frac {1}{2}\times m\times v^2[/tex]
Also, [tex]E=\frac {h\times c}{\lambda}[/tex]
[tex]\nu_0=\frac {c}{\lambda_0}[/tex]
Applying the equation as:
[tex]\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] is the wavelength of the light being bombarded
[tex]\lambda_0[/tex] is the threshold wavelength
[tex]\frac {1}{2}\times m\times v^2[/tex] is the kinetic energy of the electron emitted.
Given, [tex]\lambda=420\ nm=420\times 10^{-9}\ m[/tex]
[tex]\lambda_0=525\ nm=525\times 10^{-9}\ m[/tex]
Thus, applying values as:
[tex]\frac {6.626\times 10^{-34}\times 3\times 10^8}{420\times 10^{-9}}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{525\times 10^{-9}}+K.E.[/tex]
[tex]K.E.=\frac {6.626\times 10^{-34}\times 3\times 10^8}{420\times 10^{-9}}-\frac {6.626\times 10^{-34}\times 3\times 10^8}{525\times 10^{-9}}[/tex]
[tex]K.E.=\frac{19.878}{10^{17}\times \:420}-\frac{19.878}{10^{17}\times \:525}[/tex]
[tex]K.E.=9.4657\times 10^{-20}\ J[/tex]
