Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
<u>Binding Energy = 2.24 eV</u>
Hey!
First, let's write the problem.

Subtract the numbers, we would do the following operation,


Add 2 to both sides.

This tells us that our final answer would be,

Thanks!
-TetraFish
<span>
Of course. Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc. That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>
Yes, an object<span> that was set in motion in the past by some force, but that is no longer being acted on by a net force, is </span>moving<span> but with </span>zero acceleration<span>, i.e. it is </span>moving<span> at constant velocity.</span>
Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :



(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :



Therefore, this is the required solution.