Question

Find the longest wavelength in Paschen series.
(Given R= 1.097 x 10⁷m⁻¹ )
(Ans : 18752 $\overset{\circ}{A}$ )

Answer 1

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➡️If The Shortest Wavelength Of H Atom In Lyman Series Is "a" Then Longest Wavelength In Balmer Series Of He+ Is a) a/4 b) 5a/9 c) 4a/9 d) 9a/5. ... Here,x is the wavelength for the last line(shortest wavelength) in lyman series, .. ...... !!!!!


✌️ I THINK IT HELPED YOU ✌️

Answer 2

Dear Student,

◆ Answer -

λ = 18752 A°

● Explanation -

Rydbergs equation for wavelength in orbit is given by -

1/λ = R (1/n1² - 1/n2²)

For longest wavelength in Paschen series, n1 = 3 & n2 = 4.

1/λ = 1.097x10⁷ (1/3² - 1/4²)

1/λ = 1.097x10⁷ (1/9 - 1/16)

1/λ = 1.097x10⁷ × 7/144

λ = 144 / (1.097x10⁷ × 7)

λ = 18.752×10⁻⁷ m

λ = 18752 A°

Hence, longest wavelength in Paschen series is 18752 A°.

Thanks dear. Hope this helps you...