The work function for aluminium is 4.125 eV. The cut-off wavelength for photoelectric effect for aluminium will be

Options

A. 420 nm
B. 350 nm
C. 300 nm
D. 200 nm

Answer

C

Explanation

The work function for aluminium,
`W_(0)=4.125eV`
`=4.125xx1.6xx10^(-19)=6.6xx10^(-19)J`
the relation for work function is given by
`W_(0)=(hc)/(lamda)` (where, `lamda` is the cut-off wavelength)
`6.6xx10^(-19)=(6.6xx10^(-34)xx3xx10^(8))/(lamda)`
`lamda=(6.6xx10^(-34)xx3xx10^(8))/(6.6xx10^(-19))`
`=(19.8xx10^(-26))/(6.6xx10^(-19))`
`=3xx10^(-7)m`
`=300xx10^(-9)m=300nm`