The maximum kinetic energy of a photoelectron is 3eV . What is its stopping potential ?

Explanation

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 to maximum 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 `