The relationship between energy of a single photon and its wavelength can be determined using the formula E=hc/lambda where E is energy, h is Planck's constant, c is the speed of light, and lambda is photons.
Before being able to solve for energy, need to convert nanometers to meters.
407 nm (1 m/1 x 10^9 nm) = 4.07 x 10^-7 m
Then plug in the values we know into the equation.
E h(Planck's constant) c(speed of light)
E = (6.63 x 10^-34 Js)(3 x 10^8 m/s) / 4.07 x 10^-7 m (lambda)
E=(0.000000000000000000000000000000000663js)(300,000,000m/s)=1.989×10^-25j/ms
E=1.989x10^-25j/ms /{divided by} 4.07x10^-7m = 4.8869779x10^-33 J (the meters cancel out)
E = 4.89 x 10^-33 J
This gives us the energy in Joules of a single photon. Now, we can find the number of photons in 0.897 J
0.897J / 4.89 x 10^-33 J = ((0.897 J) / 4.89) x ((10^(-33)) J) = 1.8343558 x 10^-34
1.83435583 × 10-34m4 kg2 / s4 photons
Answer:
efficiency of heating with this oven is 51 %
Explanation:
to raise the temp of 200 ml of coffee from 30°C to 60°C the energy input to microwave oven is:
1100 J/s x 45 = 49,500 J
AT 100% efficiency
For 1°C the energy required to raise the temperature of 1 ml = 4.2 J
So for 30 C°, 1°C the energy required to raise the temperature of 200 ml =
Q = (4.2) (200)(30) = 25,200 J
efficiency = 25,200/49,500 = 0.51 = 51%
?? Is that the whole question?
We are asked in the problem to convert teh unit mci or millicurie to another unit of radioactivity, becquerel. The conversion to be used in this case is <span>1 millicurie [mCi] = 37,000,000 becquerel [Bq] .The problem gives 15 mCi which when multiplied to </span>37,000,000 is equal to 5,550,000,000 bQ
