JEE Main 16 March 2021 Shift 1 Solved Paper

Since, we know that,
‌

1

λ

=Rz2(‌

1

n12

−‌

1

n22

)......(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
‌

1

λ1

=Rz2(‌

1

22

−‌

1

32

)
⇒‌‌‌

1

λ1

=Rz2(‌

5

36

)...(ii)
Similarly, for 3rd spectral line, n1=2 and n2=5
‌

1

λ3

=Rz2(‌

1

22

−‌

1

52

)
⇒‌‌‌

1

λ3

=Rz2(‌

21

100

)......(iii)
Now, dividing Eq. (iii) by Eq. (ii), we get
‌

‌ 1 λ3

1 λ1

=‌

Rz2( 21 100 )

Rz2(‌ 5 36 )

⇒‌

λ1

λ3

=‌

21

100

×‌

36

5

⇒‌‌‌

λ1

λ3

=1.512=15.12×10−1
Comparing with the given value in the question i.e., x×10−1, the value of x=15.