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The energy of the electron in the second...

The energy of the electron in the second and third Bohr's orbitals of the hydrogen atom is `-5.42 xx 10^(-12) erg` and `-2.42 xx 10^(-12) erg` respectively ,Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.

Text Solution

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`DeltaE = E_(3) - E_(2) = [-2.41] - (-5.42)] xx 10^(-12) ~= 3 xx 10^(-12)` ergs.
Now, `3 xx 10^(-12) " ergs" = (hc)/(lamda) = (6.626 xx 10^(-27) xx 3 xx 10^(10))/(lamda) = (1.9878 xx 10^(-16))/(lamda)`
So, `lamda = (1.9878 xx 10^(-16))/(3 xx 10^(-12)) = 6.626 xx 10^(-5)cm`
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