Chapter 28, Problem 32P

To determine

The cut off wavelength, ${\text{λ}}_{\text{0}}\text{=}\frac{\text{1240}\text{nm}}{\text{V}}$

Answer to Problem 32P

It is shown that the cut off wavelength, ${\text{λ}}_{\text{0}}\text{=}\frac{\text{1240}\text{nm}}{\text{V}}$

Explanation of Solution

Given info: Cut off wavelength, ${\text{λ}}_{\text{0}}\text{=}\text{1240}\frac{\text{nm}}{\text{V}}$

Formula used:

The energy of photon with shortest wavelength must equivalent the max kinetic energy of an electron:

  $h{f}_0=\frac{hc}{{\lambda }_0}=eV$ or ${\lambda }_0=\frac{hc}{eV}$

Where,

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

  $\text{c}$ is speed of light, $3.00\times {10}^{8}m/s$

  $e$ is charge on electron, $1eV=1.6\times {10}^{-19}J/eV$

Calculation:

Substitute the known and given values in expression ${\lambda }_0=\frac{hc}{eV}$

  ${\lambda }_0=\frac{\left(6.63\times {10}^{-34}J\cdot s\right)\left(3.00\times {10}^{8}m/s\right)\left({10}^{-9}nm/m\right)}{\left(1.60\times {10}^{-19}J/eV\right)\left(1e\right)V}$

  ${\lambda }_0=\frac{\left(1.24\times {10}^{3}V\cdot nm\right)}{V}$

  ${\text{λ}}_{\text{0}}\text{=}\frac{\text{1240}\text{nm}}{\text{V}}$

Conclusion:

The given condition is shown.