Answer:
We know that
ħf = ф + Ekmax
where
ħ = planks constant = 6.626x10^-34 J s
f = frequency of incident light = 1.3x10^15 /s (1 Hz =
1/s)
ф = work function of the cesium = 2.14 eV
Ekmax = max kinetic energy of the emmitted electron.
We distinguish that:
1 eV = 1.602x10^-19 J
So:
2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J
So,
Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J
= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J
Answer:
5.19x10^-19 J
Kinetic energy:
In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.
Answer details:
Subject: Chemistry
Level: College
Keywords:
• Energy
• Kinetic energy
• Kinetic energy of emitted electrons
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