Answer:
Approximately [tex]3.03\times 10^{-19}\; \rm J[/tex], which corresponds to a wavelength of approximately [tex]6.56 \times 10^{2}\; \rm nm[/tex] in vacuum.
Explanation:
When the electron of a hydrogen atom transits from energy level [tex]n_1[/tex] to [tex]n_2[/tex], the energy change would be:
[tex]\displaystyle \Delta E \approx 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{(n_1)}^{2}} - \frac{1}{{(n_2)}^{2}}\right)[/tex].
For the transition from [tex]n_1 =3[/tex] to [tex]n_2 = 2[/tex]:
[tex]\begin{aligned}\Delta E &\approx 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{(n_1)}^{2}} - \frac{1}{{(n_2)}^{2}}\right)\\ &= 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{3}^{2}} - \frac{1}{{2}^{2}}\right) \\ &\approx -3.02639 \times 10^{-19} \; \rm J \\ &\approx -3.03\times 10^{-19}\; \rm J\end{aligned}[/tex].
The value of [tex]\Delta E[/tex] is negative because energy is released during this transition.
Look up Planck's Constant (for finding frequency from energy) and the speed of light in vacuum (for finding wavelength from frequency.)
- Planck's Constant: [tex]h \approx 6.62607 \times 10^{-34} \; \rm J \cdot s^{-1}[/tex].
- Speed of light in vacuum: [tex]c \approx 2.99792 \times 10^{8}\; \rm m \cdot s^{-1}[/tex].
Calculate the frequency [tex]f[/tex] of photons from this transition using the Planck-Einstein relation:
[tex]E = h \cdot f[/tex].
Therefore:
[tex]\begin{aligned}f &= \frac{E}{h}\end{aligned}[/tex].
Calculate the wavelength [tex]\lambda[/tex] of these photos in vacuum:
[tex]\begin{aligned}\lambda &= \frac{c}{f} \\ &= \frac{c}{ E/ h} \\ &= \frac{h \cdot c}{E} \\ &\approx \frac{6.62607 \times 10^{-34}\; \rm J \cdot s \times 2.99792 \times 10^{8}\; \rm m \cdot s^{-1}}{3.02639 \times 10^{-19}\; \rm J} \\ &\approx 6.65 \times 10^{-7}\; \rm m \\ &= 6.65 \times 10^{-7}\; \rm m \times \frac{10^{9}\;\rm nm}{1\; \rm m} \\ &= 6.65 \times 10^{2}\; \rm nm\end{aligned}[/tex].
