Chapter 27, Problem 70A

(a)

To determine

To Find: Threshold wavelength of tin.

Answer to Problem 70A

  $\text{250 nm}$

Explanation of Solution

Given:

Threshold frequency of tin, $v_o=1\cdot 2\times {10}^{15}\text{ Hz}$

Formula used:

Threshold wavelength can be obtained by:

  ${\lambda }_o=\frac{c}{v_o}$

Here, c is the speed of light.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}{\lambda }_o=\frac{3\times {10}^8\text{ m/s}}{1\cdot 2\times {10}^{15}}\\ {\lambda }_o=2\cdot 5\times {10}^{-7}\text{m =250 nm}\end{array}$

Conclusion:

The threshold wavelength of tin is $\text{250 nm}$ .

(b)

To determine

To Find: The work function of tin.

Answer to Problem 70A

  $4\cdot 97\text{ eV}$

Explanation of Solution

Given:

Threshold frequency of tin, $v_o=1\cdot 2\times {10}^{15}\text{ Hz}$

Formula used:

Work function can be obtained by:

  $E_o=h{v}_o$

Here, h is the Planck’s constant.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}{E}_o=h{v}_o\\ E_o=6\cdot 625\times {10}^{-34}\times 1\cdot 2\times {10}^{15}\text{J}\\ E_o=7\cdot 95\times {10}^{-19}\text{J}\\ E_o\text{=}\frac{7\cdot 95\times {10}^{-19}}{1\cdot 6\times {10}^{-19}}\text{eV}\\ E_o=4.97\text{eV}\end{array}$

Conclusion:

The work function of tin is $4\cdot 97\text{ eV}$ .

(c)

To determine

To Find: The kinetic energy of the ejected electrons in eV.

Answer to Problem 70A

  $2\cdot 47\text{ eV}$

Explanation of Solution

Given:

Threshold frequency of tin, $v_o=1\cdot 2\times {10}^{15}\text{ Hz}$

The work function of tin, $E_o=4\cdot 97\text{ eV}$ .

Wavelength of incident light, $\lambda =167\text{ nm =167}\times 1{\text{0}}^{-9}\text{m}$

Formula used:

Total energy $=$ kinetic energy $+$ work function

  $\begin{array}{l}E=K.E.+E_o\\ \Rightarrow K.E=E-E_o\\ \Rightarrow K.E=\frac{hc}{\lambda }-E_o\end{array}$

Here, $\lambda$ is the wavelength of light, h is the Planck’s constant and c is the speed of light.

Calculation:

  $\begin{array}{l}\frac{hc}{{\lambda }_{incident}}=K.E.+7\cdot 95\times {10}^{-19}\text{ J}\\ \frac{6\cdot 625\times {10}^{-34}\times 3\times {10}^8}{167\times {10}^{-}{}^9\text{ m}}-7\cdot 95\times {10}^{-19}\text{ J}=K.E.\\ \frac{hc}{{\lambda }_{incident}}=K.E.+7\cdot 95\times {10}^{-19}\text{ J}\\ 11\cdot 9\times {10}^{-19}-7\cdot 95\times {10}^{-19}\text{ J}=K.E.\\ 3\cdot 95\times {10}^{-19}\text{ J=}K.E.\\ KE\text{ =}\frac{3\cdot 95\times {10}^{-19}}{1\cdot 6\times {10}^{-19}}\text{eV}\\ KE=2\cdot 47\text{ eV}\end{array}$

Conclusion:

The kinetic energy of the ejected electrons is $2\cdot 47\text{ eV}$ .