Question

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

• A. 4125 $A^o$
• B. 2062.5 $A^o$
• C. 3000 $A^o$
• D. 6000$A^o$

Answer 1

The correct option is C 3000 $A^o$

$\varphi =\frac{hc}{{\lambda }_T}$
$4.125eV=\frac{1240}{{\lambda }_T}(hc=1240eV-nm)$
${\lambda }_T=\frac{1240}{4.125}$
$=300nm$
$=3000A^{\circ }.$

So, the answer is option (C).