MHT CET 2021 Physics Solved Paper

(c) Wavelength of spectral line in hydrogen atom (Z=1) given as
‌

1

λ

=R[‌

1

n12

−‌

1

n22

]...(i) [∵Z=1]

For last line of Lyman series,
n1=1‌ and ‌n2=∞
∴ From Eq. (i), we have
‌

1

λL

=R(‌

1

12

−‌

1

∞

)=R
⇒‌‌λL=‌

1

R

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)

Similarly, for last line of Balmer series, n1=2 and n2=∞
∴ From Eq. (i), we get
‌‌

1

λB

‌‌=R(‌

1

22

−‌

1

∞

)=‌

R

4

⇒‌‌λB‌‌=‌

4

R

Hence, from Eq. (i) and Eq. (ii), we get
‌

λL

λB

‌‌=‌

1∕R

4∕R

=‌

1

4

‌‌ or ‌‌λL:λB‌‌=1:4
⇒λB:λL‌‌=4:1