Threshold wavelength ($$\lambda_{threshold}$$) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ($$\phi$$). The relationship between work function and threshold wavelength is given by:

$$\phi = \frac{hc}{\lambda_{threshold}}$$

Where:

• $$\phi$$ is the work function in electron-volts (eV)
• $$h$$ is the Planck's constant
• $$c$$ is the speed of light
• $$\lambda_{threshold}$$ is the threshold wavelength

In this problem, we are given the product $$hc = 1242 ~\mathrm{eV}\,\mathrm{nm}$$, and the work functions $$\phi_{A} = 9 ~\mathrm{eV}$$ and $$\phi_{B} = 4.5 ~\mathrm{eV}$$. Let's calculate the threshold wavelengths for metal surfaces $$\mathrm{A}$$ and $$\mathrm{B}$$:

For metal surface $$\mathrm{A}$$:

$$\lambda_{A} = \frac{1242}{\phi_{A}} = \frac{1242}{9}$$

For metal surface $$\mathrm{B}$$:

$$\lambda_{B} = \frac{1242}{\phi_{B}} = \frac{1242}{4.5}$$

Now let's calculate the difference between the threshold wavelengths ($$\Delta\lambda$$):

$$\Delta\lambda = \lambda_{B} - \lambda_{A} = \frac{1242}{4.5} - \frac{1242}{9}$$

By calculating the difference, we get:

$$\Delta\lambda = 276 - 138 = 138 ~\mathrm{nm}$$

So, the difference between the threshold wavelengths for the two metal surfaces is $$\boxed{138}\,\mathrm{nm}$$.