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When light with a wavelength of 400 nm falls on the surface of sodium, electrons with a kinetic energy of 1.05 x 105J Mol-1 are emitted.(i) What is the minimum energy needed to remove an electron from sodium?(ii) What is the maximum wavelength of light that will cause a photoelectron to be emitted? |
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Answer» E = hv = \(\frac{hc}{\lambda}\) Wavelength \(\lambda\) = 400 nm = 400 × 10−9m \(\therefore\) E = \(\frac{6.626\times10^{-34}Js\times3.0\times10^{-9}m}{400\times10^{-9}m}\) = 4.969 × 10−19 J Energy of one mole of photons = 4.969 × 10−19 J x 6.022 × 1023 mol−1 = 29.923 × 104 = 2.99 × 105 J mol−1 (i) The minimum energy needed to remove one mole of electrons from sodium = (2.99 − 1.05) × 105 J mol−1 = 1.94 × 105 J mol −1 \(\therefore\) The minimum energy for one electron from sodium = \(\frac{1.94\times10^5Jmol^{-1}}{6.022\times10^{23}electron\,mol^{-1}}\) = 0.322 × 10−18 = 3.22 × 10−19J (ii) Maximum wavelength of light \(\lambda=\frac{hc}{E}\) = \(\frac{6.626\times10^{-34}Js\times3.0\times10^8ms^{-1}}{3.22\times10^{-19}J}\) = 6.17 × 10−17m = 617 × 10−9m = 617 nm |
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