Question

1/λ=Ry(1/n21 -1/n22)
you set n1 = 1 and taken2 greater than 1, you generate what is knownas the Lyman series.
a) Find the wavelength of the first member of this series. Thevalue of h is 1.05457 e -34 J*s; the Rydberg constantfor hydrogen is 1.09735 e7 m-1 ; the Bohr radius is5.29177 e-11 m; and the ground stae energy for hydrogen is 13.6057eV.
Answer in units of nm.
b) Consider the next three members of this series. Thewavelengths of successive members of the Lyman series approach acommon limit as n2 → [infinity]. What is thislimit?
Answer in units of nm.

Answer 1

Answer:

Explanation:

Equation for hydrogen ion spectrum is as follows .

$\frac{1}{\lambda} = R ( \frac{1}{n_1^2} -\frac{1}{n_2^2} )$

If n₁ = 1 and n₂ = 2,3,4 ...., it is called Layman series of radiation.

For first member n₁ = 1 and n₂ = 2,

R = 1.09735 x 10⁷ m⁻¹

$\frac{1}{\lambda} = 1.09735\times 10^7 ( \frac{1}{1} -\frac{1}{2^2} )$

$\frac{1}{\lambda}$ = .82301 x 10⁷

λ = 1.21505 x 10⁻⁷

= 1215.05x 10⁻¹⁰ m

1215.05 A

b ) For second member of Layman series

$\frac{1}{\lambda} = 1.09735\times 10^7 ( \frac{1}{1} -\frac{1}{3^2} )$

λ = 1.0252 x 10⁻⁷

= 1025.2x 10⁻¹⁰ m

1025.2A

For third member of Layman series

$\frac{1}{\lambda} = 1.09735\times 10^7 ( \frac{1}{1} -\frac{1}{4^2} )$

λ = .97204 x 10⁻⁷

= 972 x 10⁻¹⁰ m

= 972A

If n₂ = ∝ ( infinity)

$\frac{1}{\lambda} = 1.09735\times 10^7 ( \frac{1}{1} -0 )$

λ = .9113 x 10⁻⁷

= 911.3 x 10⁻¹⁰ m

= 911.3 A