Question

The work function of platinum is $6.35\, eV$. The threshold frequency of platinum is

• A. $15.32 \times 10^{14} \,Hz$
• B. $15.32 \times 10^{16} \,Hz$
• C. $15.32 \times 10^{19}\, Hz$
• D. $15.32 \times 10^{18} \,Hz$

Answer 1

Answer: (A)

Work-function $\phi_{0}=h v_{0}$
Then threshold frequency
$v_{0}=\frac{\phi_{0}}{h}=\frac{6.35 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}$
$=15.32 \times 10^{14}\, Hz$