Question

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

• A. $4125A^o$
• B. $3000A^o$
• C. $6000A^o$
• D. $2062A^o$

Answer 1

Answer: (B)

$3000A^o$

Since work function for a metal surface is $W=\frac{hc}{{\lambda }_0}$
Where ${\lambda }_0$ is threshold wavelength or cut-off wavelength here $W=4.125eV=4.125\times 1.6\times {10}^{-19}Joule$

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