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The electronic transitions from n = 2 to...

The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where 11 = principal quantum state)

A

`Li^(2+)`

B

`He^(+)`

C

H

D

`H^(+)`

Text Solution

Verified by Experts

`1/lamda=Z^(2)R_(H)[1/(n_(1)^(2))-1/(n_(2)^(2))]`
`n_(1)=1,n_(2)=2`
`1/lamda=Z^(2)R_(H)[1/1-1/4]rArr1/lamda=3/4R_(H)Z^(2)`
`thereforelamdaprop1/(Z^(2))`
Hence, elements having high atomic number will produce shortest wavelength.
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