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.

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Mergus Mergus · Answer 1 · 2019-11-05T05:17:58+01:00

Answer:

[tex]Binding\ energy=43.43\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

Given, [tex]\lambda=4.00\times 10^{-7}\ m[/tex]

[tex]\frac {hc}{\lambda_0}[/tex] is the binding energy or threshold energy

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

Thus, applying values as:

[tex]\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy[/tex]

[tex]\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[/tex]

[tex]\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J[/tex]

[tex]49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J[/tex]

[tex]Binding\ energy=43.43\times 10^{-20}\ J[/tex]