TS EAMCET 19-July-2022 Shift 1 Solved Paper

We know that, wavelength of spectral line in Balmer series is given as,
‌

1

λ

=RH(‌

1

n12

−‌

1

n22

), where n1=2 and n2=3,4.
⇒‌

1

λ

=RH[‌

1

22

−‌

1

n22

] ......(i)
For maximum wavelength n2=3 Form Eq. (i), we get
‌‌

1

λ‌max ‌

=RH(‌

1

22

−‌

1

32

)
‌=‌

5

36

RH
⇒‌‌λmax‌=‌

36

5RH

=‌

36

5×1×107

=7.2×10−7m
⇒‌‌λmax‌=7200×10−10m
‌=7200Å
For minimum wavelength, n2=∞
From Eq (i), we get
‌‌

1

λmin

=RH(‌

1

22

−‌

1

∞2

)=‌

RH

4

⇒λmin=‌

4

RH

=‌

4

1×107

=4×10−7m
⇒λmin=4000×10−10m=4000Å
∴‌λmax−λmin=7200−4000=3200Å