Respuesta :
The speed of light and the propagation of errors allows to find the results on the questions of the radiation emitted by the laser are:
a) The frequency is: f = 3.7 10¹⁴ Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
a) The speed of a wave is related to its wavelength and frequency.
c = λ f
[tex]f = \frac{c}{\lambda}[/tex]
Where c is the speed of light, λ the wavelength and f the frequency.
They indicate that the wavelength is λ = 800 nm = 800 10⁻⁹ m, the speed of light is a constant c = 2.99 10⁸ m/s.
f = [tex]\frac{2.99 \ 10^8}{800 \ 10^{-9}}[/tex]
F = 3.7 10¹⁴ Hz
b) Planck's equation states that the energy is proportional to the frequency of the radiation.
E = h f
Where E is the energy, h the Planck constant and f the frequency.
E = 6.63 10⁻³⁴ 3.7 10¹⁴
E = 2.46467 10⁻¹⁹ J
The uncertainty or error is the fluctuation that a magnitude may have due to the precision in the measurements, when the magnitude is calculated by some formula, the propagation of these uncertainties must be carried out.
Δm = ∑ [tex]\sum \frac{dm}{dx_i} | \Delta x_I|[/tex]
the expression for energy is:
E = [tex]\frac{hc}{\lambda }[/tex]
[tex]\Delta E = \frac{dE}{d \lambda} |D\lambda |[/tex]
[tex]\Delta E = \frac{h c }{\lambda^2 } |\Delta \lambda |[/tex]
When the error in the measured magnitude is not explicitly indicated, we assume that the error is in the last digit written, therefore
Δλ = ± 1 nm = ± 1 10⁻⁹ m
We calculate.
[tex]\Delta E = \frac{6.63 \ 10^{-34} \ 2.99 \ 10^8 }{(800 \ 10^{-9})^2} 1 \ 10^{-9}[/tex]
ΔE = 3.1 10⁻²² J
the error is given with a significant figure.
ΔE = 3 10⁻²² J = 0.004 10⁻¹⁹ J
The result of the energy is:
E = (2.465 ± 0.004) 10⁻¹⁹ J
In conclusion, using the speed of light and the propagation of errors, we can find the results on the questions of the radiation emitted by the laser are:
a) The frequency is; f = 3.7 1014 Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
Learn more here: brainly.com/question/15557220
