Question

A tellurium-sapphire laser can produce light at wavelength of 800 nm in ultrashort pulses that last only 4.00x10-15s (4.00 femtoseconds, or 4.00 fs). The energy in a single pulse produced by one such laser is 2.00 μJ = 2.00x10-6 J, and the pulses propagate in the positive x-direction. Find a) the frequency of the light; b) the energy and minimum energy uncertainty of a single photon in the pulse

Answer 1

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

           $f = \frac{c}{\lambda}$

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 = $\frac{2.99 \ 10^8}{800 \ 10^{-9}}$  

         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 = ∑   $\sum \frac{dm}{dx_i} | \Delta x_I|$  

the expression for energy is:

        E = $\frac{hc}{\lambda }$  

        $\Delta E = \frac{dE}{d \lambda} |D\lambda |$  

        $\Delta E = \frac{h c }{\lambda^2 } |\Delta \lambda |$

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.

        $\Delta E = \frac{6.63 \ 10^{-34} \ 2.99 \ 10^8 }{(800 \ 10^{-9})^2} 1 \ 10^{-9}$  

        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