Final answer:
The de Broglie wavelength of an electron in the first orbit of the Bohr model is equal to the circumference of that orbit, which can be expressed as λ = 2πr.
Step-by-step explanation:
The question asks about the de Broglie wavelength of an electron in the first orbit of a Bohr model. According to the de Broglie hypothesis, the wavelength (λ) of a particle is given by λ = h / p, where h is the Planck constant and p is the momentum of the particle. In the Bohr model, the momentum of an electron in an orbit is quantized and the circumference of the orbit is an integer multiple of the wavelength, which can be expressed as 2πr = nλ, with n being a positive integer representing the orbit number (quantum number).
For the first orbit (n = 1), the circumference of the orbit is 2πr, and it should be equal to the de Broglie wavelength (λ). Thus, when n = 1, λ = 2πr. Therefore, for the first Bohr orbit, the de Broglie wavelength of the electron equals the circumference of the orbit.