Question

For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equation 1 λ = 2π2mk2e4 h3c (Z2) 1 n2f − 1 n2i . Using this equation, show that the ranges of wavelengths in these two series overlap.

Answer 1

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

e is the charge of electron

h is the plank constant

c is the speed of light in vaccum

z is the atomic number = 1

$\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

where R is the  Rydberg constant = 1.097373 × 10⁷m⁻¹

For  Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap