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The frequency of revolution of an electron in nth orbit is `f_(n)`. If the electron makes a transition from nth orbit to `(n = 1)` th orbit , then the relation between the frequency `(v)` of emitted photon and `f_(n)` will beA. `v = f_(n)^(2)`B. `v = sqrt f_(n)`C. `v = (1)/(f_(n)`D. `v = f_(n)` |
Answer» Correct Answer - D `mr omega^(2) = (ke^(2))/(r ^(2)) or omega^(2) = (ke^(2))/(mr ^(3))` or `4 pi ^(2) f_(n)^(2) = (ke^(2))/(mr ^(3)) or f_(n)^(2) = (ke^(2))/(4 pi^(2) mr ^(3))` But, `r = (1)/(k) xx (n^(2) h^(2))/(4 pi^(2) me ^(2))` `:. f_(n)^(2) = (ke^(2)(k xx 4 pi^(2) me ^(2))^(2))/(4 pi^(2) m (n ^(2) h^(2))^(3))` or `f_(n)^(2) = (k^(4) e^(8) (4 pi^(2))^(2) m^(2))/(n^(2) h^(2))^(3)` or `f_(n) = ( 4 pi ^(2)k^(2) me^(4))/(n^(3) h ^(3))` Again , `hv = h^(2) = k^(2) ( 2 pi^(2) me^(4))/(h^(2)) [(1)/((n - 1)^(2)) - (1)/(n^(2))]` or `v = k^(2) ( 2 pi^(2) me^(4))/(h^(2)) [(n^(2) - (n - 1)^(2))/( n^(2) (n - 1)^(2))]` `= k^(2) ( 2 pi^(2) me^(4))/(h^(3)) [((2 n - 1))/( n^(2) (n - 1)^(2))]` If `n` "is very large , then" `v = k^(2) (2 pi^(2) ke^(4))/(h^(3)) xx (2 n)/(n^(4))` `= (4 pi ^(2) k^(2) me^(4))/(n^(3) h^(3)) = f_(n)` |
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