Question

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

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

Answer 1

Answer: (A)

3000 $\mathring{A}$

Answer 2

$\phi =hv= \frac {hc}{λ}$
$\Rightarrow λ= \frac {hc}{\phi}= \frac {1242 eV.nm}{4.125}\approx 3000 \mathring{A}$