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Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:
A
$\le 2.8\times10^{-12}$ m
B
$< 2.8\times10^{-10}$ m
C
$< 2.8\times10^{-9}$ m
D
$\ge 2.8\times10^{-9}$ m
Detailed Solution
Photon energy: $E = \frac{12400}{5000}$ eV = 2.48 eV
$(KE)_{max} = 2.48 - 2.28 = 0.20$ eV
$\lambda = \frac{h}{\sqrt{2mE}} \approx 28$ Å for the fastest electrons
Slower electrons have longer wavelengths, so $\lambda \ge 2.8\times10^{-9}$ m.
$(KE)_{max} = 2.48 - 2.28 = 0.20$ eV
$\lambda = \frac{h}{\sqrt{2mE}} \approx 28$ Å for the fastest electrons
Slower electrons have longer wavelengths, so $\lambda \ge 2.8\times10^{-9}$ m.