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A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy. Level the hydrogen atoms would be excited ? Calculate the wavelengths of the first memeber of Lyman and first member of Balmer series. |
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Answer» Solution :Here, `DeltaE`= 12.5 eV Energy of an electron in the nth orbit of hydrogen atom is `EN = -13.6/n^2`eV For ground state n = 1, `E_1=-13.6/1^2eV=-13.6` eV For first excited state n = 2, `E_2=-(13.6)/2^2`eV=-3.4 eV For SECOND excited state n = 3, `E_3=-13.6/3^2`eV=-1.51 eV Energy required to excite hydrogen atoms from ground state to the second excited state `=E_("final")-E_("initial")` = –151 –(–13.6) = 12.09 eV Thus hydrogen atoms would be excited upto third energy level (n = 3). For LYMAN SERIES, `1/lambda=R(1/(n^2f)-2/(n_i^2))` `1/lambda=1.097xx10^7(1/1^2-1/2^2)` `1/lambda=1.097xx10^7xx3/4` `1/lambda=0.82275xx=10^7 m^(-1)` `lambda=122xx10^(-9)`m =122 nm For Balmer series `1/lambda=1.097xx10^7(1/2^2-1/3^3)` `1/lambda=1.097xx10^7xx5/36` `1/lambda=0.15236xx10^7` `lambda`=656.3 nm |
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