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The threshold wavelength of tungsten is 2.76 × 10-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10-5 cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases (i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and(ii) radiation of frequency 4 × 1015 Hz is made incident on the tungsten surface. |
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Answer» Data: λ0 = 2.76 × 10-5 cm = 2.76 × 10-7 m, λ =1.80 × 10-5 cm = 1.80 × 10-7 m, v = 4 × 1015 Hz, h = 6.63 × 10-34 J∙s,c = 3 × 108 m/s (a) For λ > λ0 , v < v0 (threshold frequency). ∴ hv < hv0 . Hence, no photoelectrons are emitted. (b) Maximum kinetic energy of electrons ejected = hc \((\frac{1}{\lambda}-\frac{1}{\lambda_o})\) = (6.63 × 10-34 )(3 × 108) \((\frac{10^7}{1.8}-\frac{10^7}{2.76})\)J = (6.63 × 10-19) (0.5555 – 0.3623) = (6.63)(0.1932 × 10-19) J = 1.281 × 10-19 J = \(\frac{1.281\times10^{-19}J}{1.6\times10^{-19}{\frac{J}{eV}}}\) = 0.8006 eV (c) Maximum kinetic energy of electrons ejected = hv – \(\frac{hc}{\lambda_o}\) =(6.63 × 10-34 (4 × 1015 ) – \(\frac{(6.63\times10^{-34})(3\times10^8)}{2.76\times10^{-7}}\) = 26.52 × 10-19 – 7.207 × 10-19 = 19.313 × 10-19 J = \(\frac{19.313\times10^{-19}J}{1.6\times10^{-19}{\frac{J}{eV}}}\) = 12.07eV |
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