AP EAPCET 25-Aug-2021 Shift 2 Paper

According to H-spectrum;
Rydberg formula ⇒

1

λ

=RH(

1

ni2

−

1

n22

)
For first line, n1=2n2=3 (Balmer series)
λ=656‌nm

1

656

=RH[

1

22

−

1

32

]
=

5

36

RH ...(i)
For second line, n1=2 and n2=4

1

λ

=RH[

1

22

−

1

42

]
=

3

16

RH ...(ii)
Dividing Eqs. (i) and (ii) we get,

λ

656

=

5

36

×

16

3

λ=485.9‌nm
Similarly, wavelength of limiting line i.e. n2=∞ and n1=2

1

λ

=RH[

1

22

−

1

∞2

]=

RH

4

=

109737

4

∴ λ=364,4‌nm
Hence, wavelength of second line is 485.9‌nm and limiting line is 364.4‌nm.