Question

A metal was irradiated by light of frequency $3.2\times {10}^{15}{s}^{-1}$. The photoelectron produced had its KE, 2 times the KE of the photoelectron which was produced when the same metal was irradiated with a light of frequency $2.0\times {10}^{15}{s}^{-1}$. What is the ionization energy of metal?

Answer 1

Let ionisation energy be $\varphi$

$E={KE}_{max}+\varphi$

For $E=hv=\frac{6.626\times {10}^{-34}\times 3.2\times {10}^{-15}}{1.6\times {10}^{-19}}=13.252eV$

For $v=2\times {10}^{15}{s}^{-1}E=hv=\frac{6.626\times {10}^{-34}\times 2\times {10}^{15}}{1.6\times {10}^{-19}}=8.2826eV13.252={2KE}_{max}+\varphi \to \left(I\right)8.2826={KE}_{max}+\varphi \to \left(II\right)\varphi =3.3132eV$