We first calculate the energy contained in one photon of this light using Planck's equation:
E = hc/λ
E = 6.63 x 10⁻³⁴ x 3 x 10⁸ / 590 x 10⁻⁹
E = 3.37 x 10⁻²² kJ/photon
Now, one mole of atoms will excite one mole of photons. This means that 6.02 x 10²³ photons will be excited
(3.37 x 10⁻²² kJ/photon) x (6.02 x 10²³ photons / mol)
The energy released will be 202.87 kJ/mol
we are asked to multiply 5.4 mm by 6.02 mm
5.4 mm has 2 significant figures
6.02 mm has 3 significant figures
When multiplying these 2 numbers the answer should have the least number of significant figures
From the 2 numbers the least number of significant figures is 2 , therefore answer should be rounded off to 2 significant numbers
5.4 x 6.02 = 32.5 rounded off is 33
Answer is b. 33 m^2
<span>it tells you the sequence in which events occurred, not how long ago they occurred.</span>
T<span>his is a straightforward question related to the surface energy of the droplet. </span>
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
