The spectral line observed at 434 nm in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the nth orbit. What is the value of n?
[Rydberg constant, `(R_(H))=109677cm^(-1)`]

Options

A. 3
B. 4
C. 5
D. 6

Answer

C

Explanation

Given,
Rydberg constant `(R_(H))=109677cm^(-1)`
Wavelength `(lambda)` of the line in the Balmer series is = 434 nm
Atomic number (Z) of hydrogen is = 1
Integer value of `n_(1)=2`
From Rydberg equation,
`" "1/lambda=R_(H)Z^(2)(1/n_(1)^(2)-1/n_(2)^(2))`
`1/(434 times 10^(-7)cm)=(109677cm^(-1))(1)^(2)(1/(2)^(2)-1/(n_(2))^(2))`
`1/(434 times 10^(-7)cm^(-1))=(109677cm^(-1))(1)^(2)(1/4-1/n_(2)^(2))`
`23041.47=1096 77 times 1/4-109677 times 1/n_(2)^(2)`
`-4377.48=-109677 times 1/n_(2)^(2)`
`n_(2)^(2)=1/0.0399 rArr 25.06 rArr n_(2)=5`