
Определите энергию и массу фотона, длина волны которого равна 0.5мкм.помогите, пожалуйста..заранее
спасибо)

Ответы на вопрос

энергия Е=6.626*10^-34*0.5*10^-6=3.313*10^-40 Дж
у фотона нет массы покоя m=0



Energy of a Photon:
The energy of a photon can be calculated using the equation:E = hf
where: - E is the energy of the photon, - h is Planck's constant (approximately 6.626 x 10^-34 J·s), - f is the frequency of the photon.
Relationship between Frequency and Wavelength:
The frequency of a photon is related to its wavelength by the equation:c = λf
where: - c is the speed of light (approximately 3 x 10^8 m/s), - λ is the wavelength of the photon, - f is the frequency of the photon.
By rearranging this equation, we can express the frequency in terms of the wavelength:
f = c / λ
Calculating the Energy and Mass of a Photon:
To calculate the energy of a photon, we need to know its frequency. Since the wavelength of the photon is given as 0.5 μm (micrometers), we can use the equation above to find the frequency:f = c / λ
Substituting the values, we get:
f = (3 x 10^8 m/s) / (0.5 x 10^-6 m)
Simplifying, we find:
f = 6 x 10^14 Hz
Now that we have the frequency, we can calculate the energy of the photon using the equation:
E = hf
Substituting the values, we get:
E = (6.626 x 10^-34 J·s) x (6 x 10^14 Hz)
Simplifying, we find:
E = 3.9756 x 10^-19 J
To calculate the mass of a photon, we can use Einstein's mass-energy equivalence principle, which states that the energy of a photon is related to its mass by the equation:
E = mc^2
where: - m is the mass of the photon, - c is the speed of light.
Rearranging this equation, we can express the mass in terms of the energy:
m = E / c^2
Substituting the values, we get:
m = (3.9756 x 10^-19 J) / (3 x 10^8 m/s)^2
Simplifying, we find:
m = 4.4173 x 10^-36 kg
Therefore, the energy of the photon with a wavelength of 0.5 μm is approximately 3.9756 x 10^-19 J, and its mass is approximately 4.4173 x 10^-36 kg.
Please note that these calculations are based on the given values and the equations mentioned above.


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