Question

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding energy of the electron plus the kinetic energy of the emitted electron. When photons of 4.00×10-7 m light strike a calcium metal surface, electrons are ejected with a kinetic energy of 6.26×10-20 J. a Calculate the binding energy of the calcium electrons.

Answer 1

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

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

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted.  = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$