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An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If the photon emitted has a wavelength of 657 nm, what is the value of ni?

Respuesta :

znk

Answer:

[tex]\boxed{3}[/tex]

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

[tex]\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )[/tex]

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

[tex]\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}[/tex]

Data:

[tex]n_{f} = 2[/tex]

λ = 657 nm

Calculation:  

[tex]\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}[/tex]