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Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` thenA. `(16)/(lambda_(1)) = (9)/(lambda_(2))`B. `(16)/(lambda_(2)) = (3)/(lambda_(1))`C. `(4)/(lambda_(1)) = (1)/(lambda_(2))`D. `(16)/(lambda_(1)) = (3)/(lambda_(2))` |
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Answer» Correct Answer - B |
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