What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.50×1015Hz1.50×1015Hz ? Express your answer in joules to three significant figures.

Kinetic Energy = 6.51x10^-19 J (3 s.f)

Explanation:

frequency = 1.50×1015 s^-1

Kinetic Energy = ?

Kinetic Energy = ħf - ф

h = plank's constant = 6.626x10^-34 Js

hf = 6.626x10^-34 * 1.50×10^15 = 9.939x10^-19 J

Work function, ф = 2.1 eV = 3.3642x10^-19 J (upon conversion to joules

Kinetic energy = 9.939x10^-19 J - 3.3642x10^-19 J

Kinetic Energy = 6.511x10^-19 J

Kinetic Energy = 6.51x10^-19 J (3 s.f)

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.

answer: they produce more slow than unicellular organisms.