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Calculate the wavelength in angstrom of the photon that is emitted when an electron in Bohr orbit n=2 returns to the orbit n=1 in the hydrogen atom. The ionisation potential of the ground state of hydrogen atom is 2.17 xx10^(-11) erg per atom. |
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Answer» Solution :SINCE the hydrogen atom has only one orbit containing only one ELECTRON, the ionisation potential of the ground state of the hydrogen atom is the energy of the electron of the first orbit, i.e, `E_(1)= -2.17 xx 10^(-11)` erg Thus, `E_(2)= (E )/(n^(2))` `= -(2.17 xx 10^(-11))/(2^(2))` `therefore` energy of the radiation EMITTED, `DeltaE= E_(2)- E_(1)` `= - (2.17 xx 10^(-11))/(2^(2)) -(-2.17 xx 10^(-11))` `=1.627 xx 10^(-11)` erg We know `DeltaE= hv= (hc)/(LAMDA)` Thus `(hc)/(lamda)= 1.627 xx 10^(-11)` `lamda= (6.62 xx 10^(-27) xx 3 xx 10^(10))/(1.627 xx 10^(-11)) = 1.22 xx 10^(-5)cm` `=1220 Å (1 Å= 10^(-8)cm)` |
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