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In the previous parts, you saw that there is not equality between the de Broglie wavelength of an electron in the hydrogen atom and the circumference of its orbit. However, there does exist a definite relationship. What is the relationship between the circumference of the orbit of the n th energy level cn and the de Broglie wavelength λn? Express your answer in terms of n and λn.

Respuesta :

Answer:

[tex]L = n\lambda_n[/tex]

Explanation:

Length of the circumference is given as

[tex]L = 2\pi r[/tex]

here we know that

[tex]r = 0.529\frac{n^2}{z} A^o[/tex]

[tex]r = 0.529\frac{n^2}{1}A^o[/tex]

now we have

[tex]L = 2\pi(0.529 \frac{n^2}{z})A^o = 3.32 n^2 A^o[/tex]

now we also know that the speed of the electron in nth orbit is given as

[tex]v = 2.18 \times 10^6 \frac{z}{n} m/s[/tex]

now the de-Broglie wavelength is given as

[tex]\lambda_n = \frac{h}{mv}[/tex]

[tex]\lambda_n = \frac{h}{m(2.18 \times 10^6 \frac{1}{n})}[/tex]

[tex]\lambda_n = 3.32 n A^o[/tex]

now we have

[tex]\frac{L}{\lambda_n} = n[/tex]

[tex]L = n\lambda_n[/tex]

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