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The seond line of Balmer series has wavelength `4861 Å` The wavelength o fthe first line Balmer series isA. `1216 Å`B. `6563 Å`C. `4340 Å`D. `4101 Å` |
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Answer» Correct Answer - B For first member, `(1)/(lambda_(1)) = R[(1)/(2^(2))-(1)/(3^(2))]` or `(1)/(lambda_(1)) = R[(1)/(4)-(1)/(9)]` or `(1)/(lambda) = R [(9-4)/(36)]` or `lambda_(1) = (36)/(5R)` For second mumber, `(1)/(lambda) = R[(1)/(2^(2))-(1)/(4^(2))]` or `(1)/(lambda_(2)) = R[(1)/(4)-(1)/(16)]` or `(1)/(lambda_(2)) = R[(4-1)/(16)]` or `(1)/(lambda_(2)) = (3R)/(16)` or `lambda_(2) = (16)/(3R)` Now, `(lambda_(1))/(lambda_(2)) = (36)/(5R) xx (3R)/(16)` or `(lambda_(1))/(lambda_(2)) = (27)/(20)` or `lambda_(1) = (27)/(20) xx lambda_(2) = (27)/(20)xx 4861 Å = 6562.4 Å` . |
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