Question

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed of 2.2 × 106 m/s in a circular orbit of radius 5.3×10−11 m. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current I? (c) What is the magnetic moment of the atom due to the motion of the electron?

Answer 1

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

(c) $\mu=9.71*10^{-24}A\cdot m^2$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

(c) Magnetic moment is given by:

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

Answer 2

(a) The orbital period of the electron $T=1.5\times10^{-6}s$

(b) The current $I=1.1\times10^{-3}$

(c) The magnetic moment of the atom due to the motion of the electron

$\mu=3.71\times10^{-24} \frac{A}{m^2}$

What will be the orbital period, current and magnetic moment of the electron?

(a)The orbital period of the electron is the time taken by the electron to complete the single orbit. It is calculated as the distance covered by the electron divided by velocity.

$T=\dfrac{2\pi r}{v}$

$v=2.2\times10^6\frac{m}{s}$

$r=5.3\times10^{-11}m$

$T=2\pi\dfrac{\ 2.2\times10^6}{5.3\times10^{-11}}$

$T=1.5\times10^{-6}sec$

(b) The formula for current is given as

$I=\dfrac{q}{t}$

$I=\dfrac{1.6\times10^{-19}}{1.5\times 10^{-6}}$

$I=1.1\times10^{-3}A$

(c) Magnetic moment is given by:

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2$

$\mu=(1.1\times10^{-3})\times\pi \times(5.3\tines10^{-11}$

$\mu=9.71\times10^{-24}\dfrac{A}{m^2}$

Thus

(a) The orbital period of the electron $T=1.5\times10^{-6}s$

(b) The current $I=1.1\times10^{-3}$

(c) The magnetic moment of the atom due to the motion of the electron

$\mu=3.71\times10^{-24} \frac{A}{m^2}$

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