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Last line of Lyman series for H- atom ha...

Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` then

A

`(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))`

Text Solution

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The correct Answer is:
B
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