Question

If wavelength $\lambda =5400\mathring{A}$ is threshold value for a certain metal, then its work function would be:

• A. 2.3 eV
• B. 0.025 eV
• C. 10 eV
• D. 0.23 eV

Answer 1

Answer: (A)

$E=\frac{hc}{\lambda }J=\frac{.hc}{\lambda \times 1.6\times 10^{-19}}eV$
$\Rightarrow E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{5400\times 10^{-10}\times 1.6\times 10^{-19}}eV$
$\Rightarrow E=2.3eV$