1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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)

A
800 ˙A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
825 ˙A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
975 ˙A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1025 ˙A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 975 ˙AThe number of possible emission lines is given by, N=n(n−1)2 So, To get six possible emission lines, ⇒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.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Planck's Quantum Hypothesis
CHEMISTRY
Watch in App
Join BYJU'S Learning Program