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Obtainan expressionfor energy of an electronin Bohrorbit. Hence obtain the expression for its bindingenergy. A rectangular coilhaving 100 turnseach of length 1 cm and breadth0.5 cmis suspended in radialmagnetic induction0.002 T. The torsional constant ofsuspensionfiber is 2 xx 10^(-8) Nm/degree. Calculate the current sensitivity of a moving coil galvanometer. |
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Answer» SOLUTION :Energy of electronin Bohr orbit Accordingto Bohr'spostulate,electronrevolvesaround nucleusin circularorbit. As this motion has two energies. `:.` Total energypossessedby an electronis given by Totalenergy`(E ) = P.E. +K.E."...."(i)` Consider the electronrevolvingin the `n^(th)` orbit inhydrogen nucleus . Let `m` and -e be the MASS andcharge on the electron,r be the radius of orbit and v be LINEAR SPEED of electron. The electricpotential at a distance'r' from a charg`+e` is given by : `V = 1/(4 piepsi_(0)) .(e )/r` P.E. = Potential `xx` charge of electron P.E. `= ((1)/(4piepsi_(0)). (e)/(r ))(-e)` P.E. `= (-e^(2))/(4piepsi_(0)r)` By Bohr's `1^(st)`postulate, `(mv^(2))/(r) = (1)/(4piepsi_(0)) .(e^(2))/(r^(2))` `1/2 mv^(2) = (e^(2))/(8piepsi_(0)r)` `:.` K.E. `= (e^(2))/(8piepsi_(0)r)` `:.` Substitutingthe valuesK.E. and P.E. in equation (i), `E = (-e^(2))/(4piepsi_(0)r) + (e^(2))/(8piepsi_(0)r)` `E = (e^(2))/(8repsi_(0)r)` `E = (-me^(4))/(8epsi_(0)^(2)h^(2)n^(2))` Energy of electronin `n^(th)`orbit is - `E_(n) = -((me^(4))/(8epsi_(0)^(2)h^(2))).(1)/(n^(2))` The bindingenegy of electronis the minimum energyrequiredto make it free from the nucleus. i.e, `E_(n) = -((me^(4))/(8epsi_(0)^(2)h^(2))).1/(n^(2))` `:.` It must receive energy `= + ((me^(4))/(8epsi_(0)^(2)h^(2)). (1)/(n^(2))` to make `E_(oo) = 0` Numerical: Given : `n = 100, L = 1.0 cm = 0.01m` `b = 0.5 cm = 0.005 m, B= 0.002T` `S_(i) = ?` As we know, `S_(i) = (nBA)/(c)` `S_(i) = (100 xx 0.002 xx 0.01 xx 0.005)/(2 xx10^(-8))` `S_(i) = 0.5 xx 10^(3)` div/A `S_(i) = 500` div/A |
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