How many type II supernovae are there per stellar mass formed?

I have been searching for the number of type II (core-collapse) supernovae per unit of stellar mass formed.

It is my understanding that a star must have an initial mass of at least 8 times and no more than 40-50 times the mass of the Sun in order to end its life as a supernova. I can't seem to find the answer in any papers.

• I’m voting to close this question because it belongs on Astronomy Commented Jun 21, 2023 at 4:35
• A variant of physics.stackexchange.com/q/123422. Note, the exact upper mass limit has little influence. Commented Jun 21, 2023 at 6:56
• @naturallyInconsistent astronomy and astrophysics questions are on topic here, despite the existence of the (new) astronomy.se site. Don't push away good content just because of your misunderstanding of site policies on the matter. Commented Jun 23, 2023 at 1:51

If we call the number of star born per unit mass $$\Phi(m)$$, then the answer you are looking for is that the number of supernovae per unit mass is: $$\frac{N_{\rm sn}}{M} = \frac{ \int^{m_2}_{m_1}\ \Phi\ \ dm}{\int^{m_{\rm hi}}_{m_{\rm lo}}\ m\ \Phi\ \ dm}\ ,$$ where $$m_1$$ and $$m_2$$ are the lower and upper mass limits for supernovae progranitors and $$m_{\rm lo}$$ and $$m_{\rm hi}$$ are the lower and upper mass limits for the IMF.
We could do a toy calculation assuming the Salpeter IMF with $$\Phi = A(m/M_{\odot})^{-2.3}$$ for $$m>1 M_{\odot}$$, where $$A$$ is just a normalisation constant, and a flatter $$\Phi = A (m/M_\odot)^{-1.3}$$ for $$1.0 > m/M_{\odot} > 0.08$$. Other assumed IMFs are possible, but this should do a reasonable job and is similar to that proposed by a number of authors - e.g. Kroupa (2001). We can use $$m_1 = 8M_\odot$$, $$m_2= 40M_\odot$$, $$m_{\rm lo}=0.08M_\odot$$ (the brown dwarf limit) and $$m_{\rm hi} = 100 M_{\odot}$$. The lower limits are reasonably well understood, the upper limits less so, but their exact value has little influence on the answer because of the steepness of the Salpeter IMF. $$\frac{N_{\rm sn}}{M} = \frac{ \int^{40}_{8}\ m^{-2.3}\ \ dm}{\int^{100}_{1}\ m^{-1.3} \ dm + \int^{1}_{0.08}\ m^{-0.3} \ dm }\ = \frac{1}{81}\ .$$