1. Use the relation for threshold frequency

For photoelectric emission, the threshold frequency $\nu_0$ is given by

$\phi = h\nu_0$

So,

$\nu_0 = \frac{\phi}{h}$

where:

• $\phi = 6.63\,\text{eV}$
• $h = 6.63 \times 10^{-34}\,\text{J s}$

2. Convert work function from eV to joule

Using

$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$

we get

$\phi = 6.63 \times 1.6 \times 10^{-19}$

$\phi = 10.608 \times 10^{-19}\,\text{J}$

$\phi = 1.0608 \times 10^{-18}\,\text{J}$

3. Now calculate threshold frequency

$\nu_0 = \frac{1.0608 \times 10^{-18}}{6.63 \times 10^{-34}}$

$\nu_0 = \frac{1.0608}{6.63} \times 10^{16}$

$\nu_0 \approx 0.16 \times 10^{16}$

$\nu_0 = 1.6 \times 10^{15}\,\text{Hz}$

4. Match with the options

• A: $16 \times 10^{15}\,\text{Hz}$
• B: $16 \times 10^{12}\,\text{Hz}$
• C: $1.6 \times 10^{15}\,\text{Hz}$
• D: $1.6 \times 10^{12}\,\text{Hz}$

Hence, the correct option is:

$\boxed{\text{C}}$

5. Comparison with stored correct answer

Stored correct answer: C

My derived answer is also C, so they agree.