Chapter 27, Problem 80GP

To determine

The shortest wavelength.

Answer to Problem 80GP

The shortest wavelength is $\text{0}\text{.88}\text{nm}$ .

Explanation of Solution

Given info:

Work function is $\text{2}\text{.2}\text{eV}$ and wavelength of light is $\text{300}\text{nm}\text{= 300}\times {\text{10}}^{\text{-9}}\text{m}$ .

Formula used:

The maximum kinetic energy of the photoelectrons:

  ${\text{KE}}_{\text{max}}\text{=}\frac{\text{hc}}{\text{λ}}{\text{-W}}_{\text{0}}$

Where,

  $K{E}_{\max }$ is maximum kinetic energy.

  $\lambda$ is wavelength.

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $c$ is speed of light, ${\text{3×10}}^{\text{8}}\text{ m/s }$

  $W_0$ is work function

Where wavelength $\text{λ}$ and $\text{h}$ plank constant, $\text{c}$ speed of light , $\text{E}$ energy of different transition states.

Calculation:

Substituting the given values in the formula ${\text{KE}}_{\text{max}}\text{=}\frac{\text{hc}}{\text{λ}}{\text{-W}}_{\text{0}}$

  ${\text{KE}}_{\text{max}}\text{=}\frac{\left(\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{{\text{300×10}}^{\text{-9}}\text{m}}\text{-}\left(\text{2}\text{.2}\text{eV}\right)\left(\text{1}{\text{.60×10}}^{\text{-19}}\text{J/eV}\right)$

  ${\text{KE}}_{\text{max}}\text{=}\text{3}\text{.11}\text{×}{\text{10}}^{\text{-19}}\text{J}$

It is known that $\text{p=mv}$ and $\text{KE}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{,}\text{p}\text{=}\sqrt{\text{2}\text{m}\left(\text{KE}\right)}$ .

So,

  $\begin{array}{c}{\text{λ}}_{\text{min}}\text{=}\frac{\text{h}}{\sqrt{\text{2m}\left({\text{KE}}_{\text{max}}\right)}}\\ \text{=}\frac{\text{6}\text{.63}{\text{×10}}^{\text{-34}}\text{J}\cdot \text{s}}{\sqrt{\text{2}\left(\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)\left(\text{3}{\text{.11×10}}^{\text{-19}}\text{J}\right)}}\\ \text{=8}\text{.8}{\text{×10}}^{\text{-10}}\text{m}\\ \text{ = 0}\text{.88}\text{nm}\end{array}$

Conclusion:

The required wavelength is $\text{0}\text{.88}\text{nm}$ .