Microwave ovens emit microwave energy with a wavelength of 13.0 cm. what is the energy of exactly one photon of this microwave r

Question

TEA [102]

3 years ago

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Microwave ovens emit microwave energy with a wavelength of 13.0 cm. what is the energy of exactly one photon of this microwave r

adiation?

Physics

2 answers:

zysi [14] · Answer 1 · 2021-12-11T22:08:16+03:00

First, let's calculate the frequency of this radiation, which is given by:
$f= \frac{c}{\lambda}$
where c is the speed of light and $\lambda$ is the photon wavelength. For this radiation, photons have wavelength of
$\lambda=13.0 cm=0.13 m$
Therefore their frequency is
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{0.13 m}=2.3\cdot 10^9 Hz$

The energy of a photon with frequency f is given by
$E=hf$
where h is the Planck constant. By using the frequency we found before, we find the energy of a single photon of this radiation:
$E=hf=(6.6 \cdot 10^{-34} Js)(2.3 \cdot 10^9 Hz)=1.52 \cdot 10^{-24} J$

elena-14-01-66 [18.8K] · Answer 2 · 2021-12-11T23:54:42+03:00

The energy of exactly one photon of this microwave radiation is 1.53 × 10⁻²⁴Joule

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

$\large {\boxed {E = h \times f}}$

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$