Star Zeta Puppis lies at a distance $460$ pc from Earth. Its apparent visual magnitude $m_V$ is $2.25$, its absolute bolometric magnitude $M_{\mathrm{bol}}$ is $-9.9$, and its angular diameter is $4.3 \times 10^{−4}$ arcseconds.
(You may take for the Sun $M_{\mathrm{bol}} = +4.8$)
a) Calculate the absolute visual magnitude $M_V$ of Zeta Puppis.
b) Calculate the luminosity of Zeta Puppis in solar units $L_{\bigodot}$
For part a) I find that $M_V=m_V-5\log_{10} d +5=2.25-5\log_{10} 460 + 5\approx-6.06$
This is the correct answer.
For part b) I make use of the relations
$$F=\frac{L}{4 \pi^2 d^2}\tag{1}$$
$$m_{\zeta}-m_{\bigodot}=-2.5\log_{10}\left(\frac{F_{\zeta}}{F_{\bigodot}}\right)\tag{2}$$
Where $L$ is the luminosity, $d$ is the distance from the star to earth in parsecs, $F_{\zeta}$, $F_{\bigodot}$ are the fluxes received at Earth from Zeta Puppis and the Sun respectively. I already know that the apparent magnitude of the Sun, $m_{\bigodot}=-26.7$ (as seen from the Earth) from an earlier part of the problem sheet.
From $(1)$ and $(2)$, then
$$2.25--26.7=-2.5\log_{10}\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{d_{\bigodot}}{d_{\zeta}}\right)^2\right]\tag{3}$$
Since the distances are in parsecs $$d_{\bigodot}=\frac{1.5\times 10^{11}\mathrm{m}\,\mathrm{pc}}{3.1\times 10^{16}\mathrm{m}}\approx 4.84\times 10^{-6}\mathrm{pc}$$
So substituting given values into $(3)$ and simplifying
$$28.95=-2.5\log_{10}\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{4.84\times 10^{-6}}{460}\right)^2\right]$$
$$\implies 10^{-\dfrac{28.95}{2.5}}=\left[\frac{L_{\zeta}}{L_{\bigodot}}\left(\frac{4.84\times 10^{-6}}{460}\right)^2\right]$$
$$\implies \left(\frac{460}{4.84\times 10^{-6}}\right)^2\times10^{-11.58}=\frac{L_{\zeta}}{L_{\bigodot}}$$
$$\implies L_{\zeta}= \left(\frac{460}{4.84\times 10^{-6}}\right)^2\times10^{-11.58}L_{\bigodot}\approx \color{red}{2.38 \times 10^4 L_{\bigodot}}$$
The problem is that the correct answer is $7.6 \times 10^5 L_{\bigodot}$
The author's solution states that
To calculate the luminosity, we need to use the bolometric absolute magnitude as only bolometric quantities account for the power emitted over the whole wavelength range. As $M = −2.5 \log L+c$, where $c$ is a constant, we can relate $\zeta$ Pup’s luminosity $L_{\zeta}$ to the solar luminosity $L_{\bigodot}$ $$M_{\zeta}-M_{\bigodot}=-2.5\log_{10}\left(\frac{L_{\zeta}}{L_{\bigodot}}\right)$$ and thus $$\frac{L_{\zeta}}{L_{\bigodot}}=10^{-4 \left(M_{\zeta}-M_{\bigodot}\right)}=10^{0.4(4.8+9.9)}=7.6\times 10^5$$
Okay, so I understand why the author's solution is correct. But I don't understand why my solution is incorrect. This is because from my lecture notes I have that
Bolometric Magnitude
The total magnitude, $\color{green}{\text{apparent or absolute}}$, of a star represents the flux of the star summed over all wavelengths. This is termed the bolometric magnitude, $m_{\mathrm{bol}}$ or $M_{\mathrm{bol}}$ for $\color{green}{\text{apparent or absolute}}$. The difference between the bolometric magnitude of a star and its magnitude in a given passband, for example $V$, is called the bolometric correction, $BC$. For a given stellar type and luminosity class you can go from the magnitude measured in a given passband, say $V$, to the bolometric magnitude by adding the bolometric correction. Thus $$m_{\mathrm{bol}} = m_V + BC$$
So from the above, it clearly states that apparent magnitudes can be used also (which is what I used in my solution) but the author uses absolute magnitudes.
I don't really understand what I am missing here (probably something straightforward). So put simply, why is my solution wrong?
