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In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å : (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Åseparation ? |
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Answer» Solution :If zero of POTENTIAL energy is taken at INFINITY, the potential energy of hydrogen atom system will be `u = (q_1q_2)/(4pi epsi_0r)` In present CASE `q_1 = +1.60 xx 10^(-19)C and R =0.53 Å = 5.3 xx 10^(-11) m` `:. u = ((1.60xx10^(-19))(-1.60 xx 10^(-19))xx9xx10^9)/(5.3xx 10^(-11))J=-((1.60xx10^(-19))^2xx9xx10^9)/(5.3 xx 10^(-11) xx 1.60 xx 10^(-19))eV = -27.2 eV` (b) K.E. of electron in its orbit `K =basu/2`[Because K.E. is always positive] `:. K = 27.2/2 = 13. 6 eV` `:.` Total energy of electron `= K + u= 13.6 - 27.2 = -13.6 eV` `:.` Minimum WORK required to free the electron = + 13.6 eV (c) If zero of potential energy is taken at `r_1 = 1.06 Å = 1.06 xx 10^(-10)m`, then the potential energy of the system for `r_2 = 0.53Å = 5.3 xx 10^(-11)m` is given by `u. = (q_1q_2)/(4pi epsi_0)[1/r_2- 1/r_1] J = (q_1q_2)/(4pi epsi_0e) [ 1/r_2-1/r_1]eV` `=((+ 1.60 xx 10^(-19))(-1.60 xx10^(-19))xx9xx10^9)/(1.60 xx 10^(-19))[1/(5.3xx10^(-11))-1/(1.06xx10^(-10))]eV` `=(13.6 - 27.2)eV = -13.6 eV` but kinetic energy of electron `K. = K =+13.6 eV` `:.` Total energy of electron `=K. + u. = 13.6 = 13.6 = 0` As an amount 13.6 eV [as calculated in case (b) above] has been used up in increasing the potential energy from - 27.2 eV to - 13.6 eV. Therefore, minimum work requred to free the electron = 0 - (13.6)eV = 13.6 eV. |
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