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A photon of wavelength ` 4 xx 10^(-7) m` strikes on metal surface , the work function fo the metal being ` 2. 13 eV` Calculate : (i) the energy of the photon (ev) (ii) the kinetic energy fo the emission and the velocity fo the photoelectron `( 1 eV = 1 , 6020 xx 10^(-19) J)`, |
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Answer» Energy `(E)=(hc)/lambda=(6.6xx10^(-34)J sxx3xx10^(8) ms^(-1))/(4xx10^(-7)m)` `=4.97xx10^(-19) J=3.10 eV` b. `lambda=4xx10^(-7)m` `W_(0)=2.13 eV` `(KE=hv-W_(0)) or (hv=hv_(0)+KE)` `KE=(hc)/lambda-W_(0)=3.10 eV-2.13 eV` `KE=0.97 eV =0.97xx1.6xx10^(-19) J` c. `KE=1/2mv^(2)=0.97xx1.6xx10^(-19)J` `=1/2xx9.1xx10^(-31) kgxxv^(2)=0.97xx1.6xx10^(-19)J` `v=5.8xx10^(5) m s^(-1)` |
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