Question

What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the binding energy is 4.73 eV? Is this in the visible range?

Answer 1

Answer:

The longest wavelength is  $\lambda = 263\ nm$

It is not in the range of visible light

Explanation:

From the question the question we are told that

   The binding energy is $E = 4.73 \ eV$

the kinetic energy of the ejected photoelectron  is mathematically represented as

            $KE = hf - E$

Where h is the plank constant with a values of  $h = 4.14 *10^{-15}eV \cdot s$

         

            f is the frequency of the the EM which is mathematically represented as

     $f = \frac{c}{\lambda }$

Here c is the speed of light with value   $c = 3.0 *10^{8} m/s$

         $\lambda$ is the wavelength

 

So we have  

            $KE = h[\frac{c}{\lambda} ] - E$

So making $\lambda$ the subject of the formula

           $\lambda = \frac{hc}{[KE +E ]}$

Here the maximum kinetic energy is  zero this is because  out of all the electron ejected using a light of a threshold frequency (i.e a photo electron ) the one that has the maximum kinetic energy is none so this implies that maximum  kinetic energy is  zero so the equation becomes

           $\lambda = \frac{4.4 *10^{-15} * 3.00 *10^{8}}{(0 + 4.73)}$

           $\lambda = 263\ nm$

Looking at this we see that it is not in the range of visible light which is  

          400nm - 700nm