Question

Consider a planetary nebula with distance 1.5 kpc and angular size 12 arcsec:

a. What is its linear radius?

b. Suppose this hot *T = 30,000 K) star emits ionizing photons at a rate of 6.92*10^47 s^(-1). Assuming a recombination coefficient α = equals 3*10^(-13) cm^3/s, estimate the density of electrons in the nebula.

c. Assuming this is a spherical nebula, calculate the emission measure.

Answer 1

a) If the linear radius is r and D is the distance to the star then angular size is

The angular size is 15 arc seconds, which in radians is given as

The distance D = 1.5 kpc = 1.5*1000*3.086*10¹⁶ = 4.629*10¹⁹ m

Answer:

b) Recombination rate R is given by

The number of ionizinig photons will be the recombination rate multiplied by the volume. n_e is the density of electrons. Thus

Answer:

c) Emission measure the electron density squared multiplied by the volume