Question

When radiation of $\lambda =253.7\text{nm}$ strikes the copper plate, the energy required to stop the ejection of electrons from copper plate is $0.24\text{eV}$. Calculate the work function of copper.

• A. $4.64\text{eV}$
• B. $6.65\text{eV}$
• C. $8.45\text{eV}$
• D. $10.25\text{eV}$

Answer 1

The correct option is A $4.64\text{eV}$

Given: $\lambda =253.7\text{nm}$
$K.E{.}_{max}=0.24\text{eV}$
We know, $K.E{.}_{max}=E-\varphi$
or $\varphi =E-K.E{.}_{max}$
$E=\frac{hc}{\lambda }=\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{253.7\times {10}^{-9}}$
$=7.8\times {10}^{-19}\text{J}=4.88\text{eV}$
Therefore, $\varphi =4.88eV-0.24\text{eV}=4.64\text{eV}$