TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Wavelength of spectral lines of Hydrogen like atom is given as
‌

1

λ

=R(‌

1

n12

−‌

1

n22

)
For Balmer series, n1=2 and n2=3,4,5,6,.....
∴‌‌‌

1

λ

=R(‌

1

22

−‌

1

n22

) ...(i)
For minimum wavelength, n2=∞
From Eq. (i),
‌

1

λmin

=R(‌

1

22

−‌

1

∞2

)
‌

1

λmin

=‌

R

4

⇒λmin=‌

4

R

...(ii)
For maximum wavelength, n2=3
∴ From Eq. (i), we have

∴ From Eq. (ii) and (iii), we have
‌

λmax

λmin

=‌

36/5R

4/R

=‌

36

20

=‌

9

5