Question

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Answer 1

Answer:

Note It Down First...

Name NI

Lyman 1

Balmer 2

Paschen 3

Brackets 4

P-Fund 5

Humphrey 6

$\frac{1}{ \lambda} = rh \: \{\frac{1}{ {n}^{2} } - \frac{1}{ {n}^{2}} \} \\ \\ \frac{1}{900 \times 1 {0}^{7} } = {10}^{5}cm \{ \frac{1}{ {n}^{2} - \frac{1}{ { \infty }^{2} } } \}$

$\frac{10^{5} }{9} = 10 ( \frac{1}{ n_{1} \: ^{2} } - 0 ) \\ \\ n_{1} \: ^{2} = 9 \\ \\ n_{1} = 3 \: paschen \: series\\ \\ \infty \to3 \: is \: answer$

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Answer 2

Using the Rydberg formula, the spectral line of H - atom is suitable for this purpose is Paschen, ∞ → 3.

• Using the Rydberg formula;

1/λ = RH(1/nf^2 - 1/ni^2)

Given that;

λ = wavelength

RH = Rydberg constant

nf = final state

ni = initial state

• When final state = 3 and initial state = ∞

Then;

1/λ =  1 × 10^7 m-1 (1/3^2 - 1/ ∞^2)

1/λ =  1 × 10^7 m-1 (1/3^2 )

λ = 900 nm

Hence, the correct answer is Paschen, ∞ → 3

Learn more about the Rydberg formula; brainly.com/question/17753747