Chapter 27, Problem 84GP

(a)

To determine

Voltage applied to bring the current to zero.

Answer to Problem 84GP

Voltage applied is $V=0.65\text{ V}$

Explanation of Solution

Given:

Wavelength, $\lambda =424\text{ nm}=424\times {10}^{-9}\text{m}$

Work function, $w_0=2.28\text{ eV}$

Speed of light, $c=3\times {10}^8\text{m/s}$

Planck’s constant, $h=6.626\times {10}^{-34}\text{Js}$

  $1\text{ eV}=1.6\times {10}^{-19}\text{J}$

Formula used:

Maximum kinetic energy of an electron is given by:

  $E_{\max }=\frac{hc}{\lambda }-w_0$

Calculation:

From the above expression,

  $\begin{array}{l}{E}_{\max }=\frac{hc}{\lambda }-w_0\\ eV=\frac{hc}{\lambda }-w_0\\ eV=\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(424\times {10}^{-9}\text{m}\right)}-\left(2.28\text{ eV}\right)\\ eV=\frac{\left(0.04688\times {10}^{-17}\text{J}\right)}{\left(1.6\times {10}^{-19}\text{J/eV}\right)}-\left(2.28\text{ eV}\right)\\ eV=0.65\text{ eV}\\ V=0.65\text{ V}\end{array}$

Conclusion:

Voltage applied is $V=0.65\text{ V}$

(b)

To determine

Maximum speed of the emitted electrons.

Answer to Problem 84GP

Maximum speed of the emitted electrons is $v_{\max }=4.78\times {10}^5\text{m/s}$

Explanation of Solution

Given:

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

  $1\text{ eV}=1.6\times {10}^{-19}\text{J}$

Formula used:

Kinetic energy is given by:

  $K{E}_{\max }=\frac{1}{2}m{v}_{\max }^2$

Calculation:

From part (a),

  $\begin{array}{l}eV=0.65\text{ eV}\\ K{E}_{\max }=0.65\text{ eV}\\ K{E}_{\max }=0.65\left(1.6\times {10}^{-19}\text{J}\right)\\ K{E}_{\max }=1.04\times {10}^{-19}\text{J}\end{array}$

Using the formula for kinetic energy

  $\begin{array}{l}\frac{1}{2}m{v}_{\max }^2=1.04\times {10}^{-19}\text{J}\\ v_{\max }=\sqrt{\frac{2\left(1.04\times {10}^{-19}\text{J}\right)}{9.1\times {10}^{-31}\text{kg}}}\\ v_{\max }=4.78\times {10}^5\text{m/s}\end{array}$

Conclusion:

Hence, the maximum speed of the emitted electrons is $v_{\max }=4.78\times {10}^5\text{m/s}$

(c)

To determine

De-Broglie wavelength of the electrons.

Answer to Problem 84GP

De-Broglie wavelength of the electrons is $\lambda =1.52\text{ nm}$

Explanation of Solution

Given:

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

Planck’s constant, $h=6.626\times {10}^{-34}\text{Js}$

Formula used:

De-Broglie wavelength is given by:

  $\lambda =\frac{h}{mv}$

Calculation:

Substituting all the values in the above expression,

  $\begin{array}{l}\lambda =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}{\left(9.1\times {10}^{-31}\text{kg}\right)\left(4.78\times {10}^5\text{m/s}\right)}\\ \lambda =1.52\times {10}^{-9}\text{m}\\ \lambda =1.52\text{ nm}\end{array}$

Conclusion:

Thus, the De-Broglie wavelength of the electrons is $\lambda =1.52\text{ nm}$