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The maximum kinetic energy of a photoelectron is 3eV . What is its stopping potential ? |
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Answer» Solution :Free energy`U = 1/2 . (q^2)/(C ) "" q = 150 mu C , U = 6 mu J` ` therefore C = (q^2)/(2U) = ( (150 xx 10^(-6) )^2)/(2 xx 6 xx 10^(-6) ) = 1.875 xx 10^(-3) F` For a spring , energy ` = 1//2 kx^2` For a capacitor energy ` = 1//2 =(q^2)/( C)` Hence x corresponds to q and spring CONSTANT k corresponds to `1/C` `k = (1)/(1.875 xx 10^(-3) ) = 533 N//m` `K.E = 1//2 mv^2` and energy of inductor` = 1//2 LI^2` Hence m corresponds to L and v corresponds to L` therefore m= 1kg ` since L = 1 henry Maximum displacement corresponds tomaximum charge ` = 150 xx 10^(-6) m` Maximum VELOCITY corresponds to maximum current ` therefore v_m iff I` `I= Q OMEGA = (Q)/(sqrt(LC) ) = (150 xx 10^(-6) )/(sqrt(1 xx 1.875 xx 10^(-3) ) ) = 3.46 xx 10^(-5) A` ` therefore v_m = 3.46 xx 10^(-3) m//s ` |
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