91Ó°ÊÓ

The wavelengths

The expression for the photoelectric effect is given follows.

$hf=\mathrm{Φ}+K_{\max }$ …(¾±)

Here, f is the incident frequency, $\mathrm{Φ}$is the work function and $K_{\max }$is the maximum kinetic energy.

The express to calculate the incident frequency is given as follows.

$f=\frac{c}{\mathrm{λ}}$

Here, $\mathrm{λ}$ is the wavelength and c is the speed of the light.

The expression to calculate the maximum kinetic energy of the electron is given as follows,

$K_{\max }=eV$

Here, e is the elementary charge and V is the stopping potential.

Derive the expression for the $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.$.

Substitute eVs for $K_{max}$, and $\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.$ forfinto equation (i).

localid="1663060320209" $$\begin{gathered} \frac{hc}{\mathrm{λ}}=\mathrm{Ï•}+eV \\ \frac{1}{\mathrm{λ}}=\frac{\mathrm{Ï•}}{hc}+\frac{1}{hc}\mathrm{eV} \end{gathered}$$ …(¾±¾±)

Calculate the value of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.$to draw the graph between $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{λ}$}\right.$ and V.

Calculate the value for 433.9 nm

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}& =& \frac{1}{433.9\mathrm{nm}}\\ & =& 0.002305{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 404.7 nm

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}& =& \frac{1}{404.7\mathrm{nm}}\\ & =& 0.002471{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 365 nm

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}& =& \frac{1}{365\mathrm{nm}}\\ & =& 0.002740{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 312.5 nm

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}& =& \frac{1}{312.5\mathrm{nm}}\\ & =& 0.0032{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 253.5 nm

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}& =& \frac{1}{253.5\mathrm{nm}}\\ & =& 0.003945{\mathrm{nm}}^{-1}\end{array}$

Determine the table as,

Determine the equation of the straight line,

$\begin{array}{rcl}\frac{1}{\mathrm{λ}}-0.002305& =& \frac{0.003945-0.002305}{2.57-0.55}\left(V-0.55\right)\\ \frac{1}{\mathrm{λ}}-0.002305& =& \frac{0.00164}{2.02}\left(V-0.55\right)\\ \frac{1}{\mathrm{λ}}& =& 0.0081V-0.00044+0.002305\\ \frac{1}{\mathrm{λ}}& =& \left(0.0081V+0.0019\right){\text{nm}}^{-1}\end{array}$ …(¾±¾±¾±)

Compare the equation (ii) and (iii)

$\begin{array}{rcl}0.0081×{10}^9{\text{m}}^{-1}& =& \frac{eV}{hc}\\ h& =& \frac{1}{0.0081×{10}^9{\text{m}}^{-1}×c}eV\end{array}$

Substitute $3×{10}^8\mathrm{m}/\mathrm{s}$ for c into above equation.

$\begin{array}{rcl}h& =& \frac{1}{0.0081×{10}^9{\mathrm{m}}^{-1}×3×{10}^8\mathrm{m}/\mathrm{s}}\mathrm{eV}\\ & =& \frac{1}{0.0243×{10}^{17}}\mathrm{eVs}\\ & =& 411.5×{10}^{-17}\\ & =& ×{10}^{-15}\mathrm{eVs}\end{array}$

Hence the value of the Plank’s constant is $4.11×{10}^{-15}\text{eVs}$.

Draw the graph between the andV.

Again, compare the equation (ii) and (iii),

$\begin{array}{rcl}\frac{\mathrm{Φ}}{hc}& =& 0.0019×{10}^9\\ \mathrm{Φ}& =& 0.0019×{10}^{9}hc\end{array}$

Substitute for hand $3×{10}^8\mathrm{m}/\mathrm{s}$ for c into above equation.

$\begin{array}{rcl}\mathrm{ϕ}& =& 0.0019×{10}^9{\mathrm{m}}^{-1}×4.11×{10}^{-15}\mathrm{eV}×3×{10}^8\mathrm{m}/\mathrm{s}\\ & =& 0.0234×{10}^2\mathrm{eV}\\ & =& 2.34\mathrm{eV}\end{array}$

Hence the value of the work function is $2.34\text{eV}$.