# Explanation for the spectrum of the sun?

My Question is, why exactely does the intensity vanish below 240 nm ? If i look at the plank's law, obviously the intensity for $\lambda \rightarrow 0$ and $\lambda \rightarrow \infty$ will vanish, but why does it already vanish for $\lambda \approx 240 nm$ ?

sincerely

• Please, I am preparing a teaching material for study in the distance and I wonder if I can use this picture (physics.stackexchange.com/questions/203411/…)? thank you State University of Rio de Janeiro, Brazil Eduardo Commented Jun 15, 2016 at 15:02
• @user3365181 as a legal matter, content on the SE network is licensed under a Creative Commons Attribution-Share Alike License, which allows people to redistribute the content under certain conditions. See the legal page for more details. Commented Jun 15, 2016 at 21:26

It seems to me that the person drawing the graph was a bit sloppy - the ideal black body radiation ("idealer Schwarzer Körper" - Temperatur 5900 K) does not cut off sharply at 240 nm as shown. Instead, it should look like this:

when calculated from Planck's Law. I suspect some bug in the method used to calculate the values in the plot you reproduced - note that at the extremes of the plot, there is the possibility of overflow if the calculation is done carelessly (or in single precision). I suspect that is what happened here.

If you want to make this graph yourself, you could use (or adapt) this Python code snippet:

import math
from scipy.constants import codata
import numpy as np
import matplotlib.pyplot as plt

D = codata.physical_constants

h = D['Planck constant'][0]
k = D['Boltzmann constant'][0]
c = D['speed of light in vacuum'][0]

pi = math.pi

def planck(T, l):
p = c*h/(k*l*T)
if (p > 700):
return 1e-99
else:
return (h*c*c)/(math.pow(l, 5.0) * (math.exp(c*h/(k*l*T))-1))

Tvec=[5900]
Lvec = np.logspace(-8, -5.3, 1000)

plt.figure()
# create a semitransparent "rainbow plot" to show where visible range is:
plt.imshow(np.tile(np.linspace(0,1,100),(2,1)), extent=[400, 800, 0, 1], aspect='auto', cmap='rainbow', alpha = 0.4)

# compute Planck for a range of wavelengths
for T in Tvec:
r = []
for l in Lvec:
r.append(planck(T, l))
plt.plot(Lvec*1e9, r/np.max(r),label='T=%d'%T)
plt.xlabel('lambda (nm)')
plt.title('wavelength distribution of black body (T=5900)')
plt.xlim((0,2500))
plt.show()

• By the way, how does one explain the area between 150 and 500 nm? Or better asked, if i reduce my wavelength from 500, why does the intensity suddenly drop so fast? What is the physical background ?
– user85397
Commented Aug 30, 2015 at 20:42
• @Mareck That is just a result of the math... which in turn is because the QM solution does not allow a high occupancy of the oscillators with the highest energy (avoiding the "UV catastrophe" was one of the early triumphs of quantum mechanics). Commented Aug 30, 2015 at 20:44
• Stars, including our own sun, are not ideal black body radiators. Some are close, but different types of stars at different stages in their life cycle obviously emit different distributions of EMR. Please see Lawrence Hugh Aller (1991). Atoms, stars, and nebulae (3rd ed.). Cambridge University Press. p. 61. ISBN 978-0-521-31040-6. Commented Jun 16, 2020 at 12:07
• See also this graphic at wikipedia Commented Jun 16, 2020 at 12:10

I don't think you can draw such conclusions from a simple graphic. They've probably just cut it off at somewhere around 200nm because the power output at shorter wavelengths is almost irrelevant.

If you want a more accurate graph for shorter wavelengths, try this one. You can set it up for 5900K and graph values below 240nm.

From the wiki article on the sun's radiation:

Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power.