Question

A clean metal surface is irradiated with light of three different wavelengths: λ1\lambda_1, λ2\lambda_2, and λ3\lambda_3. The kinetic energies of the ejected electrons are as follows: λ1\lambda_1: 2.9×1020  J2.9\times10^{-20}\;\mathrm{J}; λ2\lambda_2: approximately zero; λ3\lambda_3: 4.2×1019  J4.2\times10^{-19}\;\mathrm{J}. Arrange the light in order of increasing wavelength.

(a) λ1<λ2<λ3\lambda_1<\lambda_2<\lambda_3
(b) λ2<λ1<λ3\lambda_2<\lambda_1<\lambda_3 (c) λ3<λ2<λ1\lambda_3<\lambda_2<\lambda_1 (d) λ3<λ1<λ2\lambda_3<\lambda_1<\lambda_2 (e) λ2<λ3<λ1\lambda_2<\lambda_3<\lambda_1

Solution

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The energy possessed by a photon is given by the equation E=hν=hcλE=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:

λ3<λ1<λ2\mathrm{\lambda_3}<\mathrm{\lambda_1}<\mathrm{\lambda_2}

Therefore, D must be the correct answer.

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