Calcium has a work function of \(19 eV\) with wavelength of \(150 nm\). Calculate the maximum energy of a photo electron emitted. (leV \(=1.6 \times 10^{-19} J , h=6.6 \times 10^{-34} J\) s)
A. \(6.35 EV\) B. \(8.25 eV\) C. \(14.60 eV\) D. \(2.30 eV\)
Correct Answer: B
Explanation
For the wavelength \(=150 nm\) \(=150 \times 10^{-9} m , C =3 \times 10^{8} m / s\) \(E=h f\) \(=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{-8}}{150 \times 10^{-19}}\) we convert this to \(eV\) \(1 eV =1.6 \times 10-19 J\) \(x=1.32 \times 10^{-18} J\) \(=\frac{1.32 \times 10^{-18}}{1.6 \times 10^{-19}}=8.25 eV\) Since \(h f=8.25 eV\) and this is less than \(19 eV\) the work function, (Wo), no electron is emitted