Question

For the line spectra experiment you analyzed the Balmer series to determine ni and nf in the Rydberg equation. The Balmer series occurs in the visible region of the electromagnetic spectrum. Another set of spectral lines occurs outside of the visible region. These lines occur at 1875 nm, 1282 nm, 1094 nm, and 1005 nm. nf for this series is 3. What is ni for the spectral line at 1094 nm? Hint: It is an integer less than 10.

Answer 1

Answer:

$n_i$ = 6

Explanation:

Using the Rydberg formula as:

$\frac {1}{\lambda}=R_H\times (\frac {1}{n_{f}^2}-\frac {1}{n_{i}^2})$

where,

λ is wavelength of photon

R = Rydberg's constant (1.097 × 10⁷ m⁻¹)

n₁ is the initial final level and n₂ is the final energy level

Given that:-

$n_f$ = 3

Wavelength = 1094 nm

Also,  

1 nm = 10⁻⁹ m

So,  

Wavelength = 1094 × 10⁻⁹ m

Applying in the formula as:

$\frac{1}{1094\times 10^{-9}}=1.097\times 10^7\times (\frac{1}{3^2}-\frac{1}{n_{i}^2})$

Solving for n₁ , we get

$\left(\frac{1}{3^2}-\frac{1}{n_{i}^2}\right)=\frac{1}{1094\times \:10^{-9}\times 1.097\times \:\:10^7}$

$\frac{1}{n_{i}^2}=\frac{1}{3^2}-\left(\frac{1}{1094\times \:\:10^{-9}\times \:1.097\times \:\:\:10^7}\right)$

$\frac{1}{n_i^2}=\frac{1}{9}-\frac{100}{1200.118}$

Solving, we get that:-

$n_i$ = 6

Answer 2

The ni for the spectral line at 1094 nm is 6.

From the Rydberg equation;

1/λ = R(1/nf^2 - 1/ni^2)

λ = wavelength

nf = final energy level

ni = initial energy level

R = Rydberg constant = 1.097 × 10^7 m-1

Recall that nf = 3 then in this case,  λ = 1094 nm

1/1094 × 10^-9 = 1.097 × 10^7 (1/3^2 - 1/ni^2)

0.0833 = 0.111 -  1/ni^2

1/ni^2 =  0.111 - 0.0833

ni = 6

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