A proton is located at a distance of 0.430 m from a point charge of +8.30 µC. The repulsive electric force moves the proton until it is at a distance of 1.55 m from the charge. Suppose that the electric potential energy lost by the system is carried off by a photon that is emitted during the process. What is its wavelength?
ΔPE = PE1 – PE2 = k•q•e•(1/r1 – 1/r2) =
=9•10^9•8.3•10^-6•1.6•10^-19•(1/0.43 – 1/1.55) =
=2.01•10^-14 J.
Energy of photon
E= h•c/λ,
where h is Planck constant, c is the speed of light.
λ = h•c/E = 6.626•10^-34•3•10^8/2.01•10^-14 =9.8•10^-12 m
Well, it seems like we have a proton doing some distance gymnastics near a point charge. Let's see if we can figure out the wavelength of the photon that jumps in to steal some electric potential energy!
First, let's find the change in electric potential energy (ΔPE) of the system. To do this, we can use the formula ΔPE = q * ΔV, where q is the charge and ΔV is the change in electric potential.
Given that the charge of the proton is +e (the elementary charge) and the change in electric potential is ΔV, we have ΔPE = (+e) * ΔV.
Now, let's find the electric potential at each position. Using the formula for electric potential, V = k * q / r (where k is the Coulomb constant and r is the distance from the point charge), we can calculate the initial electric potential (Vi) when the proton is at 0.430 m and the final electric potential (Vf) when the proton is at 1.55 m.
Using the formula, Vi = k * q / ri and Vf = k * q / rf, we can substitute these values into the formula for ΔV to get ΔPE = (+e) * (Vf - Vi).
Now, we need to use the formula for the change in electric potential to find the change in electric potential energy. ΔV = Vf - Vi = (k * q / rf) - (k * q / ri).
Substituting this into the formula for ΔPE, we have ΔPE = (+e) * [(k * q / rf) - (k * q / ri)].
Since we know that electric potential energy is carried off by a photon, we can use the equation E = h * c / λ (where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength) to find the wavelength.
Rearranging the equation, we have λ = h * c / E.
Now, we just need to substitute the value of ΔPE for E and we can calculate the wavelength.
Voila! Now we have the wavelength of the photon that decided to join in the proton's repulsive adventure!
To find the wavelength of the photon emitted during the process, we need to determine the change in electric potential energy and utilize the relationship between energy and wavelength in photons.
Given:
Distance of initial location (r1) = 0.430 m
Distance of final location (r2) = 1.55 m
Charge of the point charge (q) = +8.30 µC
First, we need to find the work done by the electric force (W) in moving the proton from r1 to r2. The work done can be calculated using the equation:
W = ΔPE
where ΔPE is the change in potential energy.
The change in electric potential energy (ΔPE) for a proton moved from one location to another in the presence of an electric field can be found using the equation:
ΔPE = q * ΔV
where q is the charge of the proton and ΔV is the change in electric potential.
The electric potential at a distance r from a point charge can be found using the equation:
V = k * (q / r)
where V is the electric potential, k is Coulomb's constant (k ≈ 9 × 10^9 N•m^2/C^2), q is the charge, and r is the distance from the point charge.
For the initial location (r1), the electric potential is:
V1 = k * (q / r1)
For the final location (r2), the electric potential is:
V2 = k * (q / r2)
Therefore, the change in electric potential (ΔV) is:
ΔV = V2 - V1
Now, let's calculate the change in electric potential:
V1 = (9 × 10^9 N•m^2/C^2) * (8.30 × 10^-6 C) / 0.430 m
V2 = (9 × 10^9 N•m^2/C^2) * (8.30 × 10^-6 C) / 1.55 m
ΔV = V2 - V1
After calculating these values, the next step is to find the work done by the electric force:
W = (8.30 × 10^-6 C) * ΔV
Finally, we can find the wavelength of the emitted photon using the relationship between energy and wavelength:
E = hc / λ
where E is energy, h is Planck's constant (h ≈ 6.63 × 10^-34 J•s), c is the speed of light (c ≈ 3 × 10^8 m/s), and λ is the wavelength.
Using the equation for work done:
W = ΔPE
Since energy is conserved, the work done by the electric force (W) is equal to the energy lost by the system. Therefore, we have:
W = ΔPE = E
Now, we can substitute this into the equation for energy and wavelength to find the wavelength of the photon emitted:
λ = hc / W
By plugging in the respective values, we will be able to determine the wavelength of the photon emitted during the process.
To find the wavelength of the photon emitted during the process, we need to make use of the energy conservation principle. The energy lost by the system is equal to the energy gained by the photon.
1. Calculate the initial electric potential energy (U_initial) of the system using the formula:
U_initial = k * (q1 * q2) / r_initial
Where:
- k is the electrostatic constant (9 x 10^9 N.m^2/C^2)
- q1 and q2 are the magnitudes of the charges (+e for proton and +8.30 µC for the point charge)
- r_initial is the initial separation distance (0.430 m)
2. Calculate the final electric potential energy (U_final) of the system using the same formula, but with the final separation distance (1.55 m).
3. The change in electric potential energy (ΔU) is given by:
ΔU = U_initial - U_final
4. Since energy is conserved, the lost energy ΔU is equal to the energy gained by the photon. The energy of a photon is given by the equation:
E = h * c / λ
Where:
- h is Planck's constant (6.626 x 10^-34 J.s)
- c is the speed of light in a vacuum (3 x 10^8 m/s)
- λ (lambda) is the wavelength of the photon
5. Equating the two energies, we have:
ΔU = E
ΔU = h * c / λ
Solving for λ, we get:
λ = h * c / ΔU
6. Substitute the values of Planck's constant (h), the speed of light (c), and ΔU into the equation to find the wavelength (λ).
That's it! By following these steps, you can determine the wavelength of the photon emitted during the process.