Chapter 27, Problem 77GP

To determine

The stopping potential energy for blue light.

Answer to Problem 77GP

  $0.64V$ .

Explanation of Solution

Given:

For ultraviolet light, $\text{e}=2.10\text{V}$

Wavelength of ultraviolet light $=290\times {10}^{-9}m$

Formula used:

It is known that: $e{V}_0=\frac{hc}{\lambda }-W_0$

Where,

  $h$ is Planck’s constant.

  $c$ is speed of light.

  $\lambda$ is wavelength.

  $V_0$ is potential energy.

  $e$ is electrical charge.

Calculation:

Substitute the value in the formula, $e{V}_0=\frac{hc}{\lambda }-W_0$

  $\left(1e\right)\left(2.10v\right)\left(1.60\times {10}^{-19}J/eV\right)=\frac{\left(6.63\times {10}^{34}J\cdot s\right)\left(3.00\times {10}^{8}m/s\right)}{290\times {10}^{-9}m}-W_0$

  $\therefore W_0=\frac{\left(6.63\times {10}^{34}J\cdot s\right)\left(3.00\times {10}^{8}m/s\right)}{290\times {10}^{-9}m}-\frac{\left(1e\right)\left(2.10v\right)\left(1.60\times {10}^{-19}J/eV\right)\left(290\times {10}^{-9}m\right)}{290\times {10}^{-9}m}$

  $\therefore W_0=3.50\times {10}^{-19}J$

Now, use $W_0=3.50\times {10}^{-19}J$ for the $440nm$ light,

  $e{V}_0=\frac{hc}{\lambda }-W_0$

  $\therefore e{V}_0=\frac{\left(6.63\times {10}^{34}J\cdot s\right)\left(3.00\times {10}^{8}m/s\right)}{440\times {10}^{-9}m}-3.50\times {10}^{-19}J$

  $\therefore e{V}_0=1.02\times {10}^{-19}J$

  $\therefore e{V}_0=\frac{1.02\times {10}^{-19}J}{1.60\times {10}^{-19}J/eV}$

  $\therefore e{V}_0=0.64eV$

Conclusion: The potential difference needed to stop an electron kinetic energy of $0.64eV$ is $0.64eV$ .