Let's solve the problem step by step: --- ## **Step 1: Find the flux density \( f_V \) (in erg cm\(^{-2}\) s\(^{-1}\) Å\(^{-1}\))** The flux density in the V band is related to the apparent magnitude by: \[ m_V = -2.5 \log_{10}\left(\frac{f_V}{f_{V,0}}\right) \] where \( f_{V,0} \) is the zero point flux density for V, which is: \[ f_{V,0} = 3.75 \times 10^{-9} \text{ erg cm}^{-2} \text{s}^{-1} \text{Å}^{-1} \] Plug in the numbers and solve for \( f_V \): \[ 7.5 = -2.5 \log_{10}\left(\frac{f_V}{3.75 \times 10^{-9}}\right) \] \[ -2.5 \log_{10}\left(\frac{f_V}{3.75 \times 10^{-9}}\right) = 7.5 \] \[ \log_{10}\left(\frac{f_V}{3.75 \times 10^{-9}}\right) = -3.0 \] \[ \frac{f_V}{3.75 \times 10^{-9}} = 10^{-3} \] \[ f_V = 3.75 \times 10^{-9} \times 10^{-3} = 3.75 \times 10^{-12} \text{ erg cm}^{-2} \text{s}^{-1} \text{Å}^{-1} \] **Explanation:** We used the definition of apparent magnitude and the standard zero-point to solve for the observed flux density in the V band. **Answer (Step 1):** \[ \boxed{f_V = 3.75 \times 10^{-12} \text{ erg cm}^{-2} \text{s}^{-1} \text{Å}^{-1}} \] --- ## **Step 2: Find the absolute V magnitude (\( M_V \)) and the distance modulus** The distance modulus is: \[ m - M = 5\log_{10}(d/\text{pc}) - 5 \] Given \( m_V = 7.5 \), \( d = 25 \) pc: \[ 7.5 - M_V = 5\log_{10}(25) - 5 \] \[ \log_{10}(25) = 1.39794 \] \[ 5 \times 1.39794 = 6.9897 \] \[ 6.9897 - 5 = 1.9897 \] \[ 7.5 - M_V = 1.9897 \] \[ M_V = 7.5 - 1.9897 = 5.51 \] **Explanation:** We used the definition of the distance modulus to find the absolute magnitude. This tells us how bright the star would appear at a standard distance of 10 pc. **Answer (Step 2):** \[ \text{Absolute V magnitude: } \boxed{M_V = 5.51} \] \[ \text{Distance modulus: } \boxed{m - M = 1.99} \] --- ## **Step 3: Find the bolometric magnitudes (\( m_{\text{bol}} \) and \( M_{\text{bol}} \)), and the luminosity \( L \) in units of \( L_\odot \)** ### **Bolometric Magnitudes** Bolometric correction \( BC = M_{\text{bol}} - M_V \) (same for apparent and absolute): \[ m_{\text{bol}} = m_V + BC = 7.5 + (-0.18) = 7.32 \] \[ M_{\text{bol}} = M_V + BC = 5.51 + (-0.18) = 5.33 \] ### **Luminosity (\( L/L_\odot \))** The absolute bolometric magnitude of the Sun is \( M_{\text{bol},\odot} = 4.74 \). \[ M_{\text{bol}} - M_{\text{bol},\odot} = -2.5 \log_{10}\left(\frac{L}{L_\odot}\right) \] \[ 5.33 - 4.74 = -2.5 \log_{10}\left(\frac{L}{L_\odot}\right) \] \[ 0.59 = -2.5 \log_{10}\left(\frac{L}{L_\odot}\right) \] \[ \log_{10}\left(\frac{L}{L_\odot}\right) = -0.59 / 2.5 = -0.236 \] \[ \frac{L}{L_\odot} = 10^{-0.236} = 0.58 \] **Explanation:** We applied the bolometric correction to get the bolometric magnitudes, then used the definition of bolometric magnitude to compare the star's luminosity to the Sun. **Answer (Step 3):** - \(\boxed{m_{\text{bol}} = 7.32}\) - \(\boxed{M_{\text{bol}} = 5.33}\) - \(\boxed{L/L_\odot = 0.58}\) --- ## **Final Summary Table** | Quantity | Value | |----------------------------|----------------------------------------------------| | Flux density \(f_V\) | \(3.75 \times 10^{-12}\) erg cm\(^{-2}\) s\(^{-1}\) Å\(^{-1}\) | | Absolute V magnitude \(M_V\)| 5.51 | | Distance modulus \(m - M\) | 1.99 | | Bolometric apparent mag \(m_{\text{bol}}\) | 7.32 | | Bolometric absolute mag \(M_{\text{bol}}\) | 5.33 | | Luminosity \(L/L_\odot\) | 0.58 |