Question

Ratio between longest wavelength of H atom in Lyman
series to the shortest wavelength in Balmer series of He+ is?
​with steps pls

Answer 1

Answer:

Wave number =R[(

n

1

1

)

2

−(

n

2

1

)

2

]=

wave length

1

So,

λ

1

=RZ

2

(

2

2

1

−

3

2

1

)

=R×4×(

36

5

)

λ

1

=

9

5

R

So, the ratio isλ=

5

9

R.