Answer:
Wien peak ( λmax ) is 107.40 nm
radius of super giant is 1.086 ×
m
Explanation:
given data
temperature 27 kK
power = 100000 times of Sun
Sun radius = 6.96 × 10^8 m
to find out
Wien peak ( λmax ) and radius of supergiant (r)
solution
we will apply here first wien law to find Wien peak that is
λmax = b / t
λmax = 2.9 ×
/ 27000 = 1.0740 ×
so Wien peak ( λmax ) is 107.40 nm
and
now we apply steafay law that is
P = σ × A ×
.........................1
and we know total power output 100000 time of Sun
so we say
4πr²s
= 100000 × 4πR²s
r² = 100000 × R²
/ 
put here value
r² = 100000 × (6.96×
)² ×
/ 
r² = 1.18132 ×
r = 1.086 ×
m
so radius of super giant is 1.086 ×
m
Answer:
Option (1), option (4) and option (5)
Explanation:
The main observations of Ernest Rutherford's experiment are given below:
1. most of the positively charged particles pass straight, it means there is an empty space in the atom.
2. Very few positively charged particles retraces their path.
So,
The positively charged particles were deflected because like charges repel, that means they are deflected by protons.
Almost all the positively charge concentrate in a very small part which is called nucleus.