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A 12.3 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 second member of Lyman series and second member of Balmer series. |
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Answer» Solution :The ENERGY of electron in the nth ORBIT of hydrogen atom is `E_n=-13.6/n^2eV` when the incident beam of energy 12.3 eV is absorbed by hydrogen atom. Let the electron jump from n = 1 to n = n level. `E=E_n-E_1` `12.3=-13.6/n^2 -(-13.6/1^2)` `RARR 12.3=13.6[1-1/n^2]` `rArr 12.3/13.6=1-1/n^2` `rArr 0.9=1-1/n^2` `rArr n^2=10 rArr` n=3 That is the hydrogen aton WOULD be excited upto second excited state. For Lyman Series `1/lambda=R[1/n_f^2-1/n_i^2]` `rArr 1/lambda=1.097xx10^7[1/1-1/9]` `rArr 1/lambda=1.097xx10^7xx8/9` `rArr lambda=9/(8xx1.097xx10^7)=1.025xx10^(-7)`=102.5 nm For Balmer Series `1/lambda=1.097xx10^7[1/4-1/16]` `rArr 1/lambda=1.097xx10^7xx3/16` `rArr 1=4.86xx10^(-7)m rArr ` 1= 486 nm |
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