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Sagot :
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.
"Your question is not complete, it seems to be missing the diagram of the emission spectrum"
the diagram of the emission spectrum has been added.
From the given chart;
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- c is the speed of light = 3 x 10⁸ m/s
- f is the frequency of the wave
- λ is the wavelength
[tex]f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz[/tex]
The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- h is Planck's constant = 6.626 x 10⁻³⁴ Js
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.
Learn more here:https://brainly.com/question/15962928
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