Question

The photoelectric work function for a metal surface is 4.125eV. The cut-off wavelength for this surface is

• A. $4125\mathring{A}$
• B. $3000\mathring{A}$
• C. $6000\mathring{A}$
• D. $2062\mathring{A}$

Answer 1

Answer: (B)

$3000\mathring{A}$

Since work function for a metal surface is $W=\frac{hc}{{\lambda }_0}$

where ${\lambda }_0$ is thershold wavelength or cut-off wavelength for a metral surface.
here $W=4.125eV=4.125\times 1.6\times {10}^{-19}Joule$

so ${\lambda }_0=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.125\times 1.6\times {10}^{-19}}=3000\mathring{A}$