An electron in an atom has a speed of 2.2 × 10^6 m/s and orbits the nucleus at a distance of 5 × 10^-11m.a)What is its centripetal acceleration? b)A neutron star of radius 17 km is found to rotate at 5 revolutions per second. What is the centripetal acceleration of a point on its equator?
a) To find the centripetal acceleration of an electron in an atom, we can use the formula:
\(a = \frac{v^2}{r}\)
where:
a is the centripetal acceleration,
v is the speed of the electron, and
r is the radius of the orbit.
Plugging in the given values:
\(v = 2.2 \times 10^6 \, \text{m/s}\)
\(r = 5 \times 10^{-11} \, \text{m}\)
\(a = \frac{(2.2 \times 10^6)^2}{5 \times 10^{-11}}\)
Now, let's calculate the answer.
\(a = \frac{(2.2 \times 10^6)^2}{5 \times 10^{-11}}\)
\(a = \frac{4.84 \times 10^{12}}{5 \times 10^{-11}}\)
\(a = 9.68 \times 10^{22} \, \text{m/s}^2\)
Therefore, the centripetal acceleration of the electron is \(9.68 \times 10^{22} \, \text{m/s}^2\).
b) To find the centripetal acceleration of a point on the equator of a rotating neutron star, we use the formula:
\(a = \frac{4 \pi^2 r}{T^2}\)
where:
a is the centripetal acceleration,
\(r\) is the radius of the rotation (equator) of the neutron star, and
\(T\) is the period of rotation (the time taken for one complete revolution).
Plugging in the given values:
\(r = 17 \, \text{km} = 17 \times 10^3 \, \text{m}\)
\(T = 1 \, \text{rev} / 5 \, \text{s} = \frac{1}{5} \, \text{rev/s}\)
\(a = \frac{4 \pi^2 \times 17 \times 10^3}{(\frac{1}{5})^2}\)
Now, let's calculate the answer.
\(a = \frac{4 \pi^2 \times 17 \times 10^3}{(\frac{1}{5})^2}\)
\(a = \frac{4 \pi^2 \times 17 \times 10^3}{\frac{1}{25}}\)
\(a = \frac{4 \pi^2 \times 17 \times 10^3 \times 25}{1}\)
\(a \approx 10.6 \times 10^6 \, \text{m/s}^2\)
Therefore, the centripetal acceleration of a point on the equator of the neutron star is approximately \(10.6 \times 10^6 \, \text{m/s}^2\).