An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 2.0 cm, and the electric field within the capacitor has a magnitude of 2.3 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

7.36×10⁻¹⁵ J

Explanation:

From the question,

Then kinetic energy of the electron = electric potential × charge of the electron

Ek = q'V Equation 1

Where Ek = kinetic Energy of the electron, q' = charge of an electron, V = Electric potential.

But,

V = E×d Equation 2

Where E = Electric Field, d = distance of separation of the plates of the capacitor

Given: E = 2.3×10⁶ V/m, d = 2.0 cm = (2/100) m = 0.02 m.

Substitute into equation 2

V = 2.3×10⁶(0.02)

V = 4.6×10⁴ V.

Constant: q' = 1.6×10⁻¹⁹ C

Substitute into equation 1

Ek = 4.6×10⁴(1.6×10⁻¹⁹)

Ek = 7.36×10⁻¹⁵ J

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m