1. Use the threshold condition for photoelectric emission

   The threshold wavelength $\lambda_0$ is related to the work function $\phi$ by

   $$\phi = \frac{hc}{\lambda_0}$$

   Hence,

   $$\lambda_0 = \frac{hc}{\phi}$$

2. Apply to sodium and copper

   For sodium:

   $$\phi_{\text{Na}} = 2.3\,\text{eV}$$

   For copper:

   $$\phi_{\text{Cu}} = 4.5\,\text{eV}$$

   Since $\lambda \propto \dfrac{1}{\phi}$,

   $$\frac{\lambda_{\text{Na}}}{\lambda_{\text{Cu}}} = \frac{\phi_{\text{Cu}}}{\phi_{\text{Na}}} = \frac{4.5}{2.3} \approx 1.96$$

3. Find the nearest ratio

   $$1.96 \approx 2$$

   So,

   $$\lambda_{\text{Na}} : \lambda_{\text{Cu}} \approx 2:1$$

4. Check options

   • A: $1:2$ ❌
   • B: $4:1$ ❌
   • C: $2:1$ ✅
   • D: $1:4$ ❌

Therefore, the correct option is C.