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Answer: 20 pm=20*10^-12 m
Explanation: To solve this problem we have to use the relationship given by:
λmin=h*c/e*ΔV= 1240/60000 eV=20 pm
this expression is related with the bremsstrahlung radiation when a flux of energetic electrons are strongly stopped hitting to a catode. The electrons give their kinetic energy to the atoms of the catode.
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A trace gas is a gas which makes up less than 1% by volume of the Earth's atmosphere, and it includes all gases except nitrogen (78.1%) and oxygen (20.9%). The most abundant trace gas at 0.934% is argon.
Answer:
T=1.384×10⁶seconds
Explanation:
Given data
p (Intensity)=1.30 kw/m²
E (Energy)=1.8×10⁹ J
A (Area)=1.00 m²
T (Time required)=?
Solution
E=PT ................eq(i)
where E is energy
P is radiation power
T is time
Radiating Power is given as
P=pA
Where p is intensity
A is Area
Put P=pA in eq(i) we get
E=pAT
T=E/pA
