Answer:
[tex](\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}[/tex]
So the two wavelength range will over lap
Explanation:
The Rydberg equation is given by
[tex]\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )[/tex]
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
[tex]\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )[/tex]
where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹
For Paschen series of H spectrum
[tex]n_f = 3[/tex]
[tex]n_i = 5,6,7 ...[/tex]
in Paschen series of H spectrum
The maximum wavelength occur for [tex]n_i = 4[/tex]
[tex]\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm[/tex]
The minimum wavelength occur for [tex]n_i[/tex] = ∞
[tex]\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm[/tex]
The brackett series of H spectrum
The maximum wavelength occur for [tex]n_i = 4[/tex]
[tex]\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm[/tex]
The minimum wavelength occur for [tex]n_i[/tex] = ∞
[tex]\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm[/tex]
[tex](\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}[/tex]
So the two wavelength range will over lap
