Question

The max wavelength in angstrom
for Li2+ ion in paschen series of its emission spectrum will be

Answer 1

The max wavelength in angstrom for Li2+ ion in paschen series of its emission spectrum will bein angstrom for Li2+ ion in paschen series of its emission spectrum will be

after Lyman's series and Balmer's series, Paschen's series is the 3rd series of electron transition. so, transition occurs from n = 3 to n = 4, 5, 6, 7, ....

To get maximum wavelength, transition occurs from n = 3 to n = 4

from Rydberg's equation,

$\frac{1}{\lambda}$ = RZ² [1/3² - 1/4²]

for Li²+ ion, Z = 3

then, $\frac{1}{\lambda}$ = R(3)² [1/3² - 1/4²]

= 9R × 7/144

= 63R/144

so, $\lambda$ = 144/63R

we know, Rydberg's constant, R = 1.097 × 10^7 m^-1

so, maximum wavelength, λ = 2.0836 × 10^-7 = 2083.6 A°