A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident radiation should be
(Take: hc=12425 eV˙A)
⇒6=n2(n−1)
⇒12=n2−n
⇒n2−n−12=0
⇒n2−4n+3n−12=0
⇒n(n−4)+3(n−4)=0
⇒(n+3)(n−4)=0
⇒n=−3,4
As, n should be positive, n=4
So, the electron has to get excited to third excited state to emit 6 emission lines.
⇒ Energy to be supplied to the hydrogen atom at the ground state to get excited to the n=4 is,
ΔE=E4−E1
⇒ΔE=−0.85 eV−(−13.6 eV)
⇒hcλ=12.75 eV
⇒λ=12425 eV˙A12.75 eV
⇒λ=975 ˙A
Hence, option (C) is correct.