Question

suppose an electron was bound to a proton, as in the hydrogen atom, but by the gravitational force rather than by the electric force.
part b
what would be energy of the first bohr orbit?
e = j

Explanation:

Step1: Recall Bohr's energy formula for gravitational case

For a gravitational Bohr model, the energy of the \(n\)-th orbit is given by \(E_n=-\frac{G^2m_p^2m_e^3}{2\hbar^2n^2}\), for \(n = 1\) (first orbit), \(E_1=-\frac{G^2m_p^2m_e^3}{2\hbar^2}\).

We know:
- \(G = 6.674\times10^{-11}\space Nm^2/kg^2\)
- \(m_p=1.673\times10^{-27}\space kg\)
- \(m_e = 9.109\times10^{-31}\space kg\)
- \(\hbar=1.055\times10^{-34}\space Js\)

Step2: Substitute values into the formula

First, calculate the numerator: \(G^2m_p^2m_e^3=(6.674\times10^{-11})^2\times(1.673\times10^{-27})^2\times(9.109\times10^{-31})^3\)

\((6.674\times10^{-11})^2=4.454\times10^{-21}\)

\((1.673\times10^{-27})^2 = 2.799\times10^{-54}\)

\((9.109\times10^{-31})^3=7.546\times10^{-91}\)

Multiply these together: \(4.454\times10^{-21}\times2.799\times10^{-54}\times7.546\times10^{-91}\)

\(4.454\times2.799\times7.546\approx4.454\times21.03\approx93.67\)

Exponent: \(-21-54 - 91=-166\), so numerator \(\approx93.67\times10^{-166}=9.367\times10^{-165}\)

Denominator: \(2\hbar^2 = 2\times(1.055\times10^{-34})^2=2\times1.113\times10^{-68}=2.226\times10^{-68}\)

Now, \(E_1=-\frac{9.367\times10^{-165}}{2.226\times10^{-68}}\approx - 4.208\times10^{-97}\space J\)

Wait, maybe I made a mistake in the formula. Alternatively, the Bohr energy for gravitational case can also be derived from the fact that in Bohr model, \(E =-\frac{K^2m_e}{2\hbar^2n^2}\), where for gravitational force, \(K = Gm_pm_e\) (analogous to \(K = ke^2\) in electric case). Wait, no, the correct formula for the energy in Bohr model is \(E_n=-\frac{(Gm_pm_e)^2m_e}{2\hbar^2n^2}\) (since the centripetal force is \(F = \frac{Gm_pm_e}{r^2}=\frac{m_ev^2}{r}\), and angular momentum \(m_evr = n\hbar\), solving for \(v\) and substituting into energy \(E=\frac{1}{2}m_ev^2-\frac{Gm_pm_e}{r}\)).

Let's re - derive:

From \(G\frac{m_pm_e}{r^2}=m_e\frac{v^2}{r}\) (centripetal force), we get \(v^2=\frac{Gm_p}{r}\)

Angular momentum: \(m_evr=n\hbar\Rightarrow v = \frac{n\hbar}{m_er}\)

Substitute \(v\) into \(v^2\): \(\frac{n^2\hbar^2}{m_e^2r^2}=\frac{Gm_p}{r}\Rightarrow r=\frac{n^2\hbar^2}{Gm_pm_e^2}\)

For \(n = 1\), \(r_1=\frac{\hbar^2}{Gm_pm_e^2}\) (which matches the previous part's result if we check, since in the first part, we found \(r = 1.20\times10^{29}\space m\), let's verify \(r_1=\frac{(1.055\times10^{-34})^2}{6.674\times10^{-11}\times1.673\times10^{-27}\times(9.109\times10^{-31})^2}\)

\((1.055\times10^{-34})^2 = 1.113\times10^{-68}\)

Denominator: \(6.674\times10^{-11}\times1.673\times10^{-27}\times8.297\times10^{-61}\)

\(6.674\times1.673\times8.297\approx6.674\times13.88\approx92.64\)

Exponent: \(-11-27 - 61=-99\), denominator \(\approx92.64\times10^{-99}=9.264\times10^{-98}\)

\(r_1=\frac{1.113\times10^{-68}}{9.264\times10^{-98}}\approx1.20\times10^{29}\space m\), which matches the first part. Good.

Now, energy \(E=\frac{1}{2}m_ev^2-\frac{Gm_pm_e}{r}\)

From \(v=\frac{\hbar}{m_er}\) (for \(n = 1\)), \(v^2=\frac{\hbar^2}{m_e^2r^2}\)

So \(\frac{1}{2}m_ev^2=\frac{\hbar^2}{2m_er^2}\)

And \(\frac{Gm_pm_e}{r}=\frac{Gm_pm_e}{\frac{\hbar^2}{Gm_pm_e^2}}=\frac{G^2m_p^2m_e^3}{\hbar^2}\) (wait, no, \(r=\frac{\hbar^2}{Gm_pm_e^2}\), so \(\frac{1}{r}=\frac{Gm_pm_e^2}{\hbar^2}\), then \(\frac{Gm_pm_e}{r}=\frac{G^2m_p^2m_e^3}{\hbar^2}\))

Then \(E=\frac{\hbar^2}{2m_er^2}-\frac{G^2m_p^2m_e^3}{\hbar^2}\)

But from \(r=\frac{\hbar^2}{Gm_pm_e^2}\), \(r^2=\frac{\hbar^4}{G^2m_p^2m_e^4}\), so \(\frac{\hbar^2}{2m_er^2}=\frac{\hbar^2G^2m_p^2m_e^4}{2m_e\hbar^4}=\frac{G^2m_p^2m_e^3}{2\hbar^2}\)

So \(E=-\frac{G^2m_p^2m_e^3}{2\hbar^2}\) (since \(\frac{G^2m_p^2m_e^3}{2\hbar^2}-\frac{G^2m_p^2m_e^3}{\hbar^2}=-\frac{G^2m_p^2m_e^3}{2\hbar^2}\))

Now, let's use the values:

\(G = 6.67\times10^{-11}\), \(m_p = 1.67\times10^{-27}\), \(m_e=9.11\times10^{-31}\), \(\hbar = 1.05\times10^{-34}\)

\(G^2=(6.67\times10^{-11})^2 = 4.45\times10^{-21}\)

\(m_p^2=(1.67\times10^{-27})^2=2.79\times10^{-54}\)

\(m_e^3=(9.11\times10^{-31})^3=7.57\times10^{-91}\)

Numerator: \(4.45\times10^{-21}\times2.79\times10^{-54}\times7.57\times10^{-91}=4.45\times2.79\times7.57\times10^{-21 - 54-91}\)

\(4.45\times2.79 = 12.42\), \(12.42\times7.57\approx94.0\), so numerator \(\approx94.0\times10^{-166}=9.4\times10^{-165}\)

Denominator: \(2\hbar^2=2\times(1.05\times10^{-34})^2=2\times1.10\times10^{-68}=2.20\times10^{-68}\)

Then \(E=-\frac{9.4\times10^{-165}}{2.20\times10^{-68}}\approx - 4.27\times10^{-97}\space J\)

A more accurate calculation:

\(G = 6.67384\times10^{-11}\)

\(m_p=1.6726219\times10^{-27}\)

\(m_e=9.10938356\times10^{-31}\)

\(\hbar=1.054571817\times10^{-34}\)

\(G^2 = (6.67384\times10^{-11})^2=4.4543\times10^{-21}\)

\(m_p^2=(1.6726219\times10^{-27})^2 = 2.7976\times10^{-54}\)

\(m_e^3=(9.10938356\times10^{-31})^3=7.5461\times10^{-91}\)

Numerator: \(4.4543\times10^{-21}\times2.7976\times10^{-54}\times7.5461\times10^{-91}\)

\(4.4543\times2.7976 = 12.46\)

\(12.46\times7.5461\approx94.0\)

Exponent: \(-21-54 - 91=-166\), so numerator \(= 94.0\times10^{-166}=9.4\times10^{-165}\)

Denominator: \(2\times(1.054571817\times10^{-34})^2=2\times1.1121\times10^{-68}=2.2242\times10^{-68}\)

\(E=-\frac{9.4\times10^{-165}}{2.2242\times10^{-68}}\approx - 4.226\times10^{-97}\space J\approx - 4.23\times10^{-97}\space J\)

Answer:

\(\boxed{-4.23\times10^{-97}}\) (approximate value, the exact value may vary slightly depending on the precision of constants used)