Question

Light of wavelength${\lambda }_1=340nm$ and ${\lambda }_2=540nm$ are incident on a metallic surface. If the ratio of the speed of the electrons ejected is $2$. The work function of the metal is.

• A. $1eV$
• B. $1.85eV$
• C. $1.5eV$
• D. $2eV$

Answer 1

The correct option is B

$1.85eV$

Step 1. Given data:

1. The wavelength of incident light${\lambda }_1=340nm=340\times {10}^{-9}m$
2. The wavelength of another incident light${\lambda }_2=540nm=540\times {10}^{-9}m$
3. If the ratio of the speed of the electrons ejected$\left(\frac{v_1}{v_2}\right)$$=2$
4. Planck's constant$h=6·6\times {10}^{-34}{m}^{2}kg{s}^{-1}$
5. Speed of light$c=3\times {10}^{8}m{s}^{-1}$

Step 2. Formula used:

$Energyofphoton=workfunction+maximumenergyofemittedelectron$

$\Rightarrow \frac{hc}{\lambda }=\varphi +\frac{1}{2}m{v}^2$

Where, $h$ is planck's constant

Step 3. Calculating the work function of metal:

The work function is the energy required to eject an electron from a metal surface, which can be calculated as follows,

$$\begin{gathered} \frac{hc}{{\lambda }_1}=\varphi +\frac{1}{2}m{v^2}_1 \\ and,\frac{hc}{{\lambda }_2}=\varphi +\frac{1}{2}m{v^2}_2 \\ as{v}_1=2v_2 \end{gathered}$$

Then from the above equations, we can write,

$\begin{array}{rcl}& \Rightarrow & \frac{{\displaystyle \frac{hc}{{\lambda }_1}}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }={\left(\frac{v_1}{v_2}\right)}^2\\ & =& 4\end{array}$

Now, by equating,

$\begin{array}{rcl}& \Rightarrow & \frac{hc}{{\lambda }_1}-\varphi =4\left(\frac{hc}{{\lambda }_2}-\varphi \right)\\ & \Rightarrow & 3\varphi =\frac{4hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}\\ & \Rightarrow & 3\varphi =hc\left(\frac{4}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)\\ & =& 6·6\times {10}^{-34}\times 3\times {10}^8\times \left(\frac{4}{540}-\frac{1}{340}\right)\times {10}^9\\ & \Rightarrow & \varphi =5·6\times {10}^{-19}J(as,1eV=1·6\times {10}^{-19}J)\\ & =& 1·842eV\\ & \simeq & 1·85eV\end{array}$

Thus, the work function of the metal is$1·85eV$.

Hence, option B is the correct answer.