The difference between threshold wavelengths for two metal surfaces $$A$$ and $$B$$ having work function $$\phi_A = 9$$ eV and $$\phi_B = 4.5$$ eV in nm is:
{Given, $$hc = 1242$$ eV nm}

Solution

Given $$\phi_A = 9$$ eV, $$\phi_B = 4.5$$ eV and $$hc = 1242$$ eV·nm, the threshold wavelength $$\lambda_0$$ is given by $$\lambda_0 = \frac{hc}{\phi}\,.$$

Substituting these values yields $$\lambda_A = \frac{1242}{9} = 138$$ nm and $$\lambda_B = \frac{1242}{4.5} = 276$$ nm.

Therefore, $$\lambda_B - \lambda_A = 276 - 138 = 138$$ nm, so the correct answer is Option D: $$138$$ nm.

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The difference between threshold wavelengths for two metal surfaces $$A$$ and $$B$$ having work function $$\phi_A = 9$$ eV and $$\phi_B = 4.5$$ eV in nm is:{Given, $$hc = 1242$$ eV nm}

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The difference between threshold wavelengths for two metal surfaces $$A$$ and $$B$$ having work function $$\phi_A = 9$$ eV and $$\phi_B = 4.5$$ eV in nm is:
{Given, $$hc = 1242$$ eV nm}