# What is the maximum kinetic energy of an ejected electron if the silver metal is irradiated with...

## Question:

What is the maximum kinetic energy of an ejected electron if the silver metal is irradiated with 228-nm light? The threshold wavelength for a silver metal surface is 267 nm.

## Maximum Kinetic Energy:

The formula for the maximum kinetic energy of the given substance with the threshold wavelength {eq}\lambda _0 {/eq} is shown below:

{eq}\displaystyle K.E._{max}=\frac{hc}{\lambda }-\frac{hc}{\lambda_0 } {/eq}, where,

• {eq}h=6.626\times 10^{-34}\ J\cdot s {/eq} is the Planck's constant.
• {eq}c=3\times 10^8\ m/s {/eq} is the speed of the light.
• {eq}\lambda {/eq} is the wavelength of the substance in meters.

We are given the following data:

• The value of the wavelength is {eq}\lambda = 228\ nm {/eq}.
• The value of the threshold wavelength of the silver metal is {eq}\lambda_0=267\ nm {/eq}.

The value of the wavelength in meters is:

{eq}\begin{align*} \displaystyle \lambda &= 228\ nm\times \frac{10^{-9}\ m}{1\ nm}\\ &= 228\times 10^{-9} \ m\\ \end{align*} {/eq}

The value of the threshold wavelength in meters is:

{eq}\begin{align*} \displaystyle \lambda_0 &= 267\ nm\times \frac{10^{-9}\ m}{1\ nm}\\ &= 267\times 10^{-9} \ m\\ \end{align*} {/eq}

Substituting the values of {eq}\lambda {/eq} and {eq}\lambda_0 {/eq} in the formula of the maximum kinetic energy, we get:

{eq}\begin{align*} \displaystyle K.E._{max}&=hc\left (\frac{1}{\lambda }-\frac{1}{\lambda_0 } \right )\\ &=\displaystyle 6.626\times 10^{-34}(3\times 10^8)\left (\frac{1}{228\times 10^{-9} }-\frac{1}{267\times 10^{-9} } \right )\\ &=\displaystyle \frac{19.878\times 10^{-26}}{10^{-9}}\left (\frac{1}{228}-\frac{1}{267} \right )\\ &=\displaystyle 19.878\times 10^{-26+9}\left (\frac{267-228}{228(267)} \right )\\ &=\displaystyle 19.878\times 10^{-17}\left (\frac{39}{60876} \right )\\ &\approx \displaystyle \boxed{0.0127\times 10^{-17}\ J}\\ \end{align*} {/eq} 