Question

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

• A. $4125\stackrel{\circ }{A}$
• B. $2062.5\stackrel{\circ }{A}$
• C. $3000\stackrel{\circ }{A}$
• D. $6000\stackrel{\circ }{A}$

Answer 1

The correct option is C

$3000\stackrel{\circ }{A}$

$W.\left(eV\right)=\frac{12375}{{\lambda }_0\left(\stackrel{\circ }{A}\right)}\Rightarrow {\lambda }_0=\frac{12375}{4.125}=3000\stackrel{\circ }{A}$