Question

The radius of the first orbit of an H atom is a0. The de Broglie wavelength of an electron in the third orbit is:

A. a0/3
B. a0
C. 3a0
D. 9a0

Answer 1

The de Broglie wavelength of an electron in the third orbit of a hydrogen atom is equal to the circumference of the first Bohr orbit (a0), as the wavelength must fit an integer number of times around the orbit. Hence, the correct answer is B. a0.

Step-by-step explanation:

The question is related to the Bohr model of the hydrogen atom in quantum mechanics. According to this model, the electrons orbit the nucleus in certain discrete orbits without radiating energy. Each of these orbits corresponds to a specific energy level and has a fixed radius. The radius of the first orbit (n=1) is known as a0, sometimes called the Bohr radius.

The de Broglie wavelength is the wavelength associated with a particle and is given by h/p, where h is Planck's constant and p is the momentum of the particle. For an electron in an orbit, its momentum is quantized due to the wave-like properties described by de Broglie's hypothesis. In the Bohr model, the angular momentum (L) of an electron in orbit is quantized and given by L = n*h/(2*pi), where n is the principal quantum number of the orbit.

For an electron in the third orbit (n=3), the circumference of the orbit must be an integer multiple of the de Broglie wavelength. The circumference for the third orbit is 2*pi*3*a0. Since the angular momentum is quantized, the de Broglie wavelength (λ) for the third orbit would be the circumference of the orbit (2*pi*3*a0) divided by the number n=3, hence λ = 2*pi*3*a0/3 = 2*pi*a0. However, we have to compare the wavelength to a0, not 2*pi*a0. The de Broglie wavelength is actually the circumference divided by the angular momentum quantization (n), and since we are given that n=3, the relationship simplifies to a single orbit circumference, which is actually 2*pi*a0. The circumference of one orbit is not just a0, it is 2*pi*a0, thus the correct answer is B. a0.

Answer 2

The de Broglie wavelength of an electron in the third orbit of a hydrogen atom is 3 times the Bohr radius (3a0), which is option C. This is because the electron's orbit radius is nine times the Bohr radius and to maintain a standing wave, the wavelength must be three times the radius.

Step-by-step explanation:

The question relates to finding the de Broglie wavelength of an electron in the third orbit of a hydrogen atom. According to the Bohr model, the radius of the electron's orbit increases as the square of the principal quantum number n. As such, the third orbit's radius is nine times that of the first orbit (n=1), which is defined as a0, the Bohr radius. The de Broglie wavelength (λ) of an electron is inversely proportional to its momentum, which decreases with increasing orbit radius. This is because the electron's speed decreases in higher orbits as per Bohr's quantization conditions.

Based on these principles, the de Broglie wavelength for an electron in the third orbit is 3 times the Bohr radius (a0) because the orbit's circumference would have to accommodate 3 wavelengths to maintain a standing wave pattern. In a formulaic sense, this would be shown as λ = 2πrn / n, and when n=3, rn = 9a0, so λ = 2π*9a0 / 3 = 3 * 2πa0, simplifying to 3a0. Therefore, the de Broglie wavelength for the third orbit's electron is option C, 3a0.