Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
What that means is the atom is so radioactive that the nucleus is unstable.
A higher frequency. I just got this answer on usa test prep.
As the scattering angle of the photon increases, the wavelength associated with the photon increases.
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Explanation:</u></h3>
The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle is 180 degree then the wavelength of the photon will become double. This is known as Compton wavelength.
When a photon undergoes collision process, the photo loses its energy and this energy is transferred to the electrons. This causes energy of the photon to decrease and thus the frequency also decreases. Thus, the wavelength of the photon will increase.