The difference between threshold wavelengths for two metal surfaces $\mathrm{A}$ and $\mathrm{B}$ having work function $\phi_{A}=9 ~\mathrm{eV}$ and $\phi_{B}=4 \cdot 5 ~\mathrm{eV}$ in $\mathrm{nm}$ is:

$\{$ Given, hc $=1242 ~\mathrm{eV} \mathrm{nm}\}$

<p>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:</p> $\phi = \frac{hc}{\lambda_{threshold}}$ <p>Where:</p> <ul> <li>$\phi$ is the work function in electron-volts (eV)</li> <li>$h$ is the Planck's constant</li> <li>$c$ is the speed of light</li> <li>$\lambda_{threshold}$ is the threshold wavelength</li> </ul> <p>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 $\mathrm{B}$:</p> <p>For metal surface $\mathrm{A}$:</p> $\lambda_{A} = \frac{1242}{\phi_{A}} = \frac{1242}{9}$ <p>For metal surface $\mathrm{B}$:</p> $\lambda_{B} = \frac{1242}{\phi_{B}} = \frac{1242}{4.5}$ <p>Now let's calculate the difference between the threshold wavelengths ($\Delta\lambda$):</p> $\Delta\lambda = \lambda_{B} - \lambda_{A} = \frac{1242}{4.5} - \frac{1242}{9}$ <p>By calculating the difference, we get:</p> $\Delta\lambda = 276 - 138 = 138 ~\mathrm{nm}$ <br/> <p>So, the difference between the threshold wavelengths for the two metal surfaces is $\boxed{138}\,\mathrm{nm}$.