Question

Work function for aluminium metal is 4.125 eV. For photo electric effect from aluminium metal, the cut off wavelength will be-

• A. 200 nm
• B. 300 nm
• C. 150 nm
• D. 420 nm

Answer 1

The correct option is B 300 nm

${\lambda }_0=\frac{hc}{w_0}$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.125\times 1.6\times {10}^{-19}}$

$=3000\times {10}^{-10}m$

${\lambda }_0=300nm$