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Calculate the angular frequency of an electron occupying the second Bohr's orbit of He^(+) ion. |
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Answer» Solution :Velocity of electron in the NTH orbit of H-like particle `(v) = (2pi ZE^(2))/(nh)` Radius of electron in the nth orbit of H-like particle `(r_(n)) = (n^(2) h^(2))/(4pi^(2) me^(2) Z)` Angular frequency `(omega) = (v)/(r_(n)) = (2pi Ze^(2))/(nh) xx (4pi^(2) me^(2) Z)/(n^(2) h^(2)) = (8pi^(3) Z^(2) me^(4))/(n^(3) h^(3))` For `He^(+), Z = 2, n = 2` (GIVEN) Also `m = 9.108 xx 10^(-28)J, h = 6.625 xx 10^(-27)` erg sec, `e = 4.803 xx 10^(-10)` esu SUBSTITUTING these values, we get `omega = (8 xx (22//7)^(3) xx (2)^(2) xx (9.108 xx 10^(-28)) xx (4.803 xx 10^(-10))^(4))/((2)^(3) xx (6.625 xx 10^(-27))^(3)) = 2.067 xx 10^(16) s^(-1)` |
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