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Photoelectron are liberated by altra voilet light of wavelength `3000Å`from a metallic surface for which the photoelectron thershold is `4000 Å` calculate de broglic wavelength of electron with maximum kinetic energy |
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Answer» KE = Quantum energy - Theshold energy = `(hc)/(lambda_(1)) - (hc)/(lambda_(2))` `= (6.626 xx 10^(-34) xx 3 xx 10^(8))/(3000 xx 10^(-10)) - (6.626 xx 10^(-34) xx 3 xx 10^(8))/(4000 xx 10^(-10)) ` `= (6.626 xx 10^(-34) xx 3 xx 10^(8))/(10^(-10)) ((1)/(3000) - (1)/(4000))` `= 1.656 xx 10^(-19) J` `1/2mv^(2)=1.656xx10^(-19)` `rArr m^(2)v^(2) = 2 xx 1.6565 xx 10^(-19) xx 9.1 xx 10^(-31)` `mv = 5.49 xx 10^(25)` `rArr lambda = (h)/(mv) = (6.626 xx 10^(-34))/(3.49 xx 10^(-25)) = 1.2 xx 10^(-9) m` |
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