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(a)Show that the ratio of velocity of an electron in the firstBohr orbit to the speed of light c is a dimensionless number. (b) Compute the velocity of electrons in ground state, first excited state and second excited state in Bohr atom model. |
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Answer» Solution :(a) The velocity of an electron in `n^(th)` orbit is `v_(n) = (H)/(2pima_(0))(Z)/(n)` where `a_(0) = (epsilon_(0)h^(2))/(pi me^(2))` = Bohr radius .Substituting for `a_(0)` in `v_(n)`, `v_(n) = (e^(2))/(2epsilon_(0)^(h))(Z)/(n)= c ((e^(2))/(2epsilon_(0)hc))(Z)/(n) = (acZ)/(n)` where c is the speed of LIGHT in free space or vaccum and its value is ` c xx 3 xx10^(8) ms^(-1)` and `alpha` is called FINE structure constant. For a hydrogen atom, Z = 1 and for the first orbit , n = 1, the ratio of velocity of electron in first orbit to the speedof light in vaccum or free space is ` alpha = ((1.6 xx 10^(-19)C)^(2))/(2 xx (8.854 xx 10^(-12)C^(2) N^(-1) m^(-2))) = ((1.6 xx 10^(-19) C)^(2))/((6.6 xx 10^(-34) Nms) xx(3 xx10^(8) ms^(-1))` `(v_(1))/(c) = alpha = (e^(2))/(2 epsilon_(0)hc)` `approx (1)/(136.9) =(1)/(137)` which is a dimensionless number `Rightarrow alpha = (1)/(137)`. (B) Using fine structure constant, the velocity of electron in can be written as`v_(n) = (acZ)/(n)` For hydrogen atom (Z = 1) the velocity of electron in `n^(th)` orbit is `v_(n) = (c)/(137)(1)/(n) = (2.19 xx 10^(6))(1)/(n) ms^(-1)` For the first orbit (ground state), the velocity of electron is `v_(n) = 2.19 xx 10^(6) ms^(-1)` For the second orbit (first excited state), the velocity of electron is `v_(2) = 1.095 xx 10^(6) ms^(-1)` For the third orbit (second excited stae), the velocity of electron is `v_(3) = 0.73 xx 10^(6) ms^(-1)` Here, `v_(1) gt v_(2) gt v_(3)` |
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