Respuesta :
Answer:
[tex]K.E.=2.48\times 10^{-19}\ 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=2000\ \dot{A}=2000\times 10^{-10}\ m[/tex]
[tex]\lambda_0=2665\ \dot{A}=2665\times 10^{-10}\ m[/tex]
Thus, applying values as:
[tex]\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}+K.E.[/tex]
[tex]K.E.=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}-\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}[/tex]
[tex]K.E.==\frac{19.878}{10^{16}\times \:2000}-\frac{19.878}{10^{16}\times \:2665}[/tex]
[tex]K.E.=2.48\times 10^{-19}\ J[/tex]
