Answer:
The kinetic energy of an electron can be related to the electric potential energy by using the conservation of energy principle. The change in kinetic energy is equal to the negative change in electric potential energy:
ΔKE = -ΔPE
The kinetic energy (KE) of an electron with mass (m) and speed (v) is given by:
KE = (1/2) * m * v^2
The electric potential energy (PE) of an electron with charge (e) in an electric field with potential difference (V) is given by:
PE = e * V
Given that the potential difference VPQ = 120 V, and assuming the charge of the electron (e) is the elementary charge (1.602 × 10^-19 C), we can write:
ΔPE = e * ΔV = e * (VP - VQ)
Now, let's equate the change in kinetic energy to the negative change in electric potential energy:
(1/2) * m * vQ^2 - (1/2) * m * vP^2 = - e * (VP - VQ)
We know that the mass of an electron (m) is approximately 9.109 × 10^-31 kg.
Given that the speed of the electron at point P (vP) is 7.0 × 10^6 m/s, and VP - VQ = 120 V, we can solve for the speed of the electron at point Q (vQ):
(1/2) * (9.109 × 10^-31 kg) * vQ^2 - (1/2) * (9.109 × 10^-31 kg) * (7.0 × 10^6 m/s)^2 = - (1.602 × 10^-19 C) * (120 V)
Now, solve for vQ:
(4.5545 × 10^-31 kg) * vQ^2 - (2.98747 × 10^-13 kg m^2/s^2) = - (1.9224 × 10^-17 C V)
vQ^2 = [(2.98747 × 10^-13 kg m^2/s^2) - (1.9224 × 10^-17 C V)] / (4.5545 × 10^-31 kg)
vQ^2 ≈ 6.96369 × 10^12 m^2/s^2
vQ ≈ √(6.96369 × 10^12 m^2/s^2)
vQ ≈ 2.64 × 10^6 m/s
The speed of the electron at point Q is approximately 2.64 × 10^6 m/s.