The angular momentum of an electron in the Bohr hydrogen atom is mur , where \(m\) is the mass of the electron, \(u,\) its velocity, and \(r,\) the radius of the Bohr orbit. The angular momentum can have only the values nh/2 \(\pi\), where \(n\) is an integer (the number of the Bohr orbit). Show that the circum frences of the various Bohr orbits are integral multiples of the de Broglie wavelengths of the electron treated as a matter wave.

Short Answer

Expert verified
The circumference for each Bohr orbit is an integral multiple of an electron's de Broglie wavelength.

Step by step solution

01

Write down the known facts

Angluar Momentum L = \(mur = \frac{nh}{2\pi}\), where m is mass of electron, u its velocity, r the radius, n is an integer denoting orbit number, and h is Planck's constant. According to de Broglie hypothesis, \(\lambda = \frac{h}{mu}\), where \(\lambda\) is the wavelength.
02

Substituting for velocity

From the angular momentum equation, the velocity can be written as \(u = \frac{nh}{2\pi m r}\). Substitute this into the wavelength formula, and we find \(\lambda = \frac{2\pi r}{n}\)
03

Relate the circumference to the wavelength.

The circumference of an orbit is given by \(2\pi r\). Thus, each orbit's circumference can be written as \(n\lambda\), meaning that each orbit's circumference is an integral multiple of the electron's de Broglie wavelength.

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