Question

If the threshold wavelength for copper is 2665 A, calculate the maximum kinetic energy of a photoelectron generated by ultraviolet light of λ 2000 A. Compute the stopping potential.

Answer 1

Answer:

$K.E.=2.48\times 10^{-19}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

$\lambda_0$ is the threshold wavelength

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted.

Given, $\lambda=2000\ \dot{A}=2000\times 10^{-10}\ m$

$\lambda_0=2665\ \dot{A}=2665\times 10^{-10}\ m$

Thus, applying values as:

$\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}+K.E.$

$K.E.=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}-\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}$

$K.E.==\frac{19.878}{10^{16}\times \:2000}-\frac{19.878}{10^{16}\times \:2665}$

$K.E.=2.48\times 10^{-19}\ J$