Chemistry, 22.10.2020 09:01 coolquezzie

Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.

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Answer from: maddie306

The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.

Using the Rydberg Equation for energy;

ΔE = -RH(1/n^2final - 1/n^2initial)

Given that;

nfinal = 3

ninitial = 7

RH = 2.18 × 10^-18 J

ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)

ΔE = - 2.18 × 10^-18(0.11 - 0.02)

ΔE = - 1.96 × 10^-19 J

For the second part;

Since the wavelength is 1000nm, we have;

λ = 1000nm

c = 3 × 10^8 m/s

f = ?

c = λf

f = c/λ

f = 3 × 10^8 m/s/1000 × 10^-9 m

f = 3 × 10^8 m/s/ 1 × 10^-6 m

f = 3 × 10^14 Hz

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Answer from: HopeBordelon4

The energy of each transition is approximately $1.98\times 10^{-19}\; \rm J$ .

The frequency of photons released in such transitions is approximately $3.00\times 10^{14}\; \rm Hz$ .

Explanation:

The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.

Let $\lambda_\text{vac}$ denote the wavelength of the photon released when measured in vacuum.Let $R_\text{H}$ denote the Rydberg constant for hydrogen. $R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}$ .Let n_1

and

denote the principal quantum number of the initial and final main energy level of that electron. (Both $n_1\!$ and $n_2\!$ should be positive integers; n_1 n_2

The Rydberg Equation gives the following relation:

$\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)$ .

Rearrange to obtain and expression for $\lambda_\text{vac}$ :

$\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}$ .

In this question, n_1 = 7 while n_2 = 3 . Therefore:

$\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}$ .

Note, that $1.0\times 10^{-6}\; \rm m$ is equivalent to $1000\; \rm nm$ . That is: $1.0\times 10^{-6}\; \rm m = 1000\; \rm nm$ .

Look up the speed of light in vacuum: $c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}$ . Calculate the frequency of this photon:

$\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}$ .

Let represent Planck constant. The energy of a photon of wavelength would be $E = h \cdot f$ .

Look up the Planck constant: $h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s$ . With a frequency of $3.00\times 10^{14}\; \rm Hz$ ( $1\; \rm Hz = 1\; \rm s^{-1}$ ,) the energy of each photon released in this transition would be:

$\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}$ .