The Sun cranks out about 4 x 1026 J/s, i.e. its radiant power pe = 4 x 10²6 W. For a square meter of area here on the Earth whose surface normal points directly to the Sun, how many watts are received? How many 550-nm photons would that correspond to? (The Earth-Sun distance is about 150x106 km.)
The Sun cranks out about 4 x 1026 J/s, i.e. its radiant power pe = 4 x 10²6 W. For a square meter of area here on the Earth whose surface normal points directly to the Sun, how many watts are received? How many 550-nm photons would that correspond to? (The Earth-Sun distance is about 150x106 km.)
Related questions
Question

Transcribed Image Text:1) The Sun cranks out about 4 x 1026 J/s, i.e. its radiant power pe = 4 x 10²6 W.
For a square meter of area here on the Earth whose surface normal points directly
to the Sun, how many watts are received? How many 550-nm photons would that
correspond to? (The Earth-Sun distance is about 150x100 km.)
Expert Solution

Step 1
The energy radiated by sun is spread over the shperical surface with sun as center. So to calculate energy incident on earth's surface, we first calculate the surface area at earth-sun distance. The steps are given below.
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
