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If a light of wavelength 330nm is incident on a metal with word function 3.55eV, the electrons are emitted. Then the wavelength of the emitted electron is ( Taken h = 6.6 xx 10^(-34) Js) |
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Answer» `lt 2.75 XX 10^(-9) m ` `K_("MAX")=(hc)/(lambda)-phi_0 = ((1240)/(330)-3.55) eV= (3.76 - 3.55) eV` `K_("max")=0.21 eV` de-Brogliewavelength of emittedelectron `lambda=(h)/(sqrt(2mKE))=(6.63 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(31) xx 0.21 xx 1.6 xx 10^(-19)))=2.668 xx 10^(-9)` m ` lambda= 2.67 xx 10^(-9) m` The two WAVELENGTH of the emittedelectron is ` lt 2.75 xx 10^(-9)`m |
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