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The radiation corresponding to 3to2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3xx10^(-4)T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to : |
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Answer» 1.6 eV `(r^(2)e^(2)B^(2))/(2M)=(mv^(2))/(2)` Now `hv=phi+1//2 mv^(2)` `1.89=phi+1//2 mv^(2)` `:.1.89-phi=(r^(2)e^(2)B^(2))/(2m) (1)/(e) eV=(r^(2)e^(2))/(2m) eV` `=(100xx10^(-6)xx1.6xx10^(-19)xx9xx10^(-8))/(2xx9.1xx10^(-31))` Work function `phi=1.89-(1.6xx9)/(2xx9.1)` `=1.89-0.79~=eV` |
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