Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
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.
The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\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}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\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}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\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}[/tex].
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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