Chapter 27, Problem 67A

(a)

To determine

To Calculate: Threshold wavelength of iron.

Answer to Problem 67A

  $264\cdot 1\text{ nm}$

Explanation of Solution

Given:

Work function of iron, $E_o=4\cdot 7\text{eV}$

Formula used:

  $E_o=\frac{hc}{{\lambda }_{threshold}}$

Calculation:

  $\begin{array}{l}\frac{hc}{{\lambda }_{threshold}}=4\cdot 7\text{eV}\\ \frac{6\cdot 62\times {10}^{-34}\times 3\times {10}^8}{{\lambda }_{threshold}}=4\cdot 7\times 1\cdot 6\times {10}^{-19}\text{ J}\\ {\lambda }_{threshold}=264\cdot 1\text{ nm}\end{array}$

Conclusion:

Threshold wavelength of the elected electrons of iron is ${\lambda }_{threshold}=264\cdot 2\text{ nm}$

(b)

To determine

To Find:The maximum kinetic energy of the ejected electrons of iron.

Answer to Problem 67A

  $5\cdot 7206\times {10}^{-19}\text{ J}=3\cdot 57\text{ eV}$

Explanation of Solution

Given:

Work function of iron, $E_o=4\cdot 7\text{eV}$

Wavelength of incident radiation, $\lambda =150\text{nm}$

Formula used:

Total energy $=$ threshold energy $+$ kinetic energy

Energy, $E=hv=\frac{hc}{\lambda }$

Calculation:

  $\frac{hc}{\lambda }=\left(4\cdot 7\times 1\cdot 6\times {10}^{-19}\right)\text{ J + }$ kinetic energy

  $\frac{6\cdot 625\times {10}^{-34}\times 3\times {10}^8\text{ m/s}}{150\times {10}^{-9}\text{ m}}=\left(4\cdot 7\times 1\cdot 6\times {10}^{-19}\right)\text{ J + }$ kinetic energy

Kinetic energy $=7\cdot 52\times {10}^{-19}\text{ J}$

Kinetic energy $\frac{=7\cdot 52\times {10}^{-19}\text{ eV}}{1\cdot 6\times {10}^{-19}}=4.7\text{eV}$

Conclusion:

The kinetic energy of the ejected electrons $=4\cdot 7\text{ eV}$ .