Question

A clean metal surface is irradiated with light of three different wavelengths: $\lambda_1$, $\lambda_2$, and $\lambda_3$. The kinetic energies of the ejected electrons are as follows: $\lambda_1$: $2.9\times10^{-20}\;\mathrm{J}$; $\lambda_2$: approximately zero; $\lambda_3$: $4.2\times10^{-19}\;\mathrm{J}$. Arrange the light in order of increasing wavelength.

(a) $\lambda_1<\lambda_2<\lambda_3$
(b) $\lambda_2<\lambda_1<\lambda_3$ (c) $\lambda_3<\lambda_2<\lambda_1$ (d) $\lambda_3<\lambda_1<\lambda_2$ (e) $\lambda_2<\lambda_3<\lambda_1$

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Step 1

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The energy possessed by a photon is given by the equation $E=h\nu=\dfrac{hc}{\lambda}$. Since the wavelength of a photon is inversely proportional to its energy, an increase in wavelength corresponds to a decrease in energy. Thus, the order of decreasing wavelength must be:

$\mathrm{\lambda_3}<\mathrm{\lambda_1}<\mathrm{\lambda_2}$

Therefore, D must be the correct answer.

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