If the sortest wavelength of h atom in lyman seeis ix x then longest wavelength in balmer series of he+ is
Answers
Answer:
Explanation:
The wave number of any series is given by;
(1/ λ ) = 109678 × [ (1/n²1}) - (1/n²2]
where 109678 is a constant, say = α and the series starts from n1 to n2;
Where x is the wavelength for the last line that has the shortest wavelength in lyman series,
Therefore, n1=1 to n = 2 = ∞
= 1/x = α (1-1) - 1/∞ = α ( 1-o) = α
For wavelength of first line of the balmer series, where n1 =2 to n2 =3
Lets say the wavelength to be found be = λ
(1/λ ) = α × [(1/2²) - (1/3²)]
= α(5/36)
Substituting the value
= (1/ λ) = (5/36)(1/x) = 5/(36x)
= λ =36x/5
Thus, the longest series is 36/5
Answer:
Explanation:
The wave number of any series is given by;
(1/ λ ) = 109678 × [ (1/n²1}) - (1/n²2]
where 109678 is a constant, say = α and the series starts from n1 to n2;
Where x is the wavelength for the last line that has the shortest wavelength in lyman series,
Therefore, n1=1 to n = 2 = ∞
= 1/x = α (1-1) - 1/∞ = α ( 1-o) = α
For wavelength of first line of the balmer series, where n1 =2 to n2 =3
Lets say the wavelength to be found be = λ
(1/λ ) = α × [(1/2²) - (1/3²)]
= α(5/36)
Substituting the value
= (1/ λ) = (5/36)(1/x) = 5/(36x)
= λ =36x/5
Thus, the longest series is 36/5