Since, we know that,
‌=Rz2(‌−‌)......(i)
where,
λ= wavelength of light emitted.
R= Rydberg's constant,
Z= atomic number,
n1= principal quantum number of lower energy level
and
‌‌n2= principal quantum number of higher energy level.
Therefore, for
1st spectral line of Balmer series,
n1=2 and
n2=3 ‌=Rz2(‌−‌) ⇒‌‌‌=Rz2(‌)...(ii)
Similarly, for 3rd spectral line,
n1=2 and
n2=5 ‌=Rz2(‌−‌) ⇒‌‌‌=Rz2(‌)......(iii)
Now, dividing Eq. (iii) by Eq. (ii), we get
‌=‌⇒‌=‌×‌ ⇒‌‌‌=1.512=15.12×10−1 Comparing with the given value in the question i.e.,
x×10−1, the value of
x=15.