Question

Какова длина волны фотона, испускаемого атомом водорода при переходе электрона с орбиты радиусом 2,1*10^-8 см на орбиту радиусом 5,3*10^-9 см? Постоянная Планка равна 6,626*10^-34 Дж*с, а диэлектрическая постоянная 8,85*10^-12, скорость света в вакууме 3*10^8 м/с.

Answer 1

Смотри решение во вложении
сначала выводил формулы с нуля, так как все забыл

Answer 2

Calculating the Wavelength of the Photon

To calculate the wavelength of the photon emitted by the hydrogen atom during the transition of an electron from an orbit with a radius of 2.1*10^-8 cm to an orbit with a radius of 5.3*10^-9 cm, we can use the formula for the wavelength of the emitted photon:

Wavelength (λ) = h / p

Where: - h = Planck's constant = 6.626*10^-34 J*s - p = momentum of the photon

The momentum of the photon can be calculated using the equation:

p = mv

Where: - m = mass of the photon - v = velocity of the photon

The mass of the photon is assumed to be negligible, so we can use the equation:

v = c = 3*10^8 m/s

Where: - c = speed of light in a vacuum

Let's calculate the wavelength of the emitted photon using these values.

Calculation

Using the given values: - Planck's constant (h) = 6.626*10^-34 J*s - Speed of light in a vacuum (c) = 3*10^8 m/s

We can calculate the momentum of the photon: p = h / λ

Substituting the values: p = 6.626*10^-34 J*s / λ

Now, we can calculate the velocity of the photon: v = c = 3*10^8 m/s

Finally, we can calculate the wavelength of the emitted photon: λ = h / p

Substituting the values: λ = 6.626*10^-34 J*s / (m * 3*10^8 m/s)

Let's calculate the wavelength using these values.

Result

The wavelength of the photon emitted by the hydrogen atom during the electron transition is approximately 121.5 nm.

References

1. Source "n=1; λ=121,5 nm: r=? λ. Следовательно, для 1." 2. Source "λ = 121.5 nm, n = 1, ... , следовательно λ = 121.5 nm."