Question #01511

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Question #01511

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Answer 1 · 2017-11-17T02:42:38

Three spectral lines
4 => 3 transition is in Infrared spectrum $(λ = 1890 n m)$ $(lamda = 1890 nm)$
4 => 2 transition is in the Visible spectrum $(λ = 488 n m)$ $(lamda=488nm)$ , and
4 => 1 transition in the Ultra Violet Spectrum $(λ = 97 n m)$ $(lamda= 97nm)$ .

Explanation:

One can calculate the wavelength of each transition using the equation
$\frac{1}{λ}$ $1/lamda$ = $R_{H}$ $R_H$ $(\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ $(1/n_f^2 - 1/n_i^2)$ ; $R_{H} = 109, 678 c m^{- 1}$ $R_H = 109,678cm^-1$ .

4 => 3 transition
=> $\frac{1}{λ}$ $1/lamda$ = $109, 678 c m^{- 1} (\frac{1}{{(3)}^{2}} - (\frac{1}{{(4)}^{2}})$ $109,678cm^-1(1/(3)^2 - (1/(4)^2)$ = $5265 c m^{- 1}$ $5265cm^-1$

=> $λ$ $lamda$ = $1.89 \times 10^{- 4} c m$ $1.89xx10^-4cm$ = 1890 nm (IR wavelength)

4 => 2 transition
=> $\frac{1}{λ}$ $1/lamda$ = $109, 678 c m^{- 1} (\frac{1}{{(2)}^{2}} - (\frac{1}{{(4)}^{2}})$ $109,678cm^-1(1/(2)^2 - (1/(4)^2)$ = $2051 c m^{- 1}$ $2051cm^-1$

=> $λ$ $lamda$ = $4.88 \times 10^{- 5} c m$ $4.88xx10^-5cm$ = 488 nm (Visible wavelength)

4 => 1 transition
=> $\frac{1}{λ}$ $1/lamda$ = $109, 678 c m^{- 1} (\frac{1}{{(1)}^{2}} - (\frac{1}{{(4)}^{2}})$ $109,678cm^-1(1/(1)^2 - (1/(4)^2)$ = $5265 c m^{- 1}$ $5265cm^-1$

=> $λ$ $lamda$ = $9.73 \times 10^{- 6} c m$ $9.73xx10^-6cm$ = 97 nm (UV wavelength)
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Question #01511

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