What potential difference is needed to stop an electron having an initial speed of 3.70x 105 m/s? Answer in volts
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To determine the potential difference needed to stop an electron with an initial speed of 3.70 × 10^5 m/s, we need to consider the relationship between electric potential energy (U) and kinetic energy (K) for a charged particle.
The electric potential energy of a charged particle in an electric field is given by:
U = qV,
where U is the electric potential energy, q is the charge of the particle, and V is the potential difference.
The kinetic energy of an electron is given by:
K = (1/2)mv^2,
where K is the kinetic energy, m is the mass of the electron, and v is its velocity.
In this case, the electron must come to a stop, meaning that its kinetic energy must be fully converted into electric potential energy.
Setting U equal to K and rearranging the equation, we have:
qV = (1/2)mv^2.
Since the charge of an electron (q) is known to be -1.6 × 10^-19 Coulombs and its mass (m) is approximately 9.11 × 10^-31 kilograms, we can substitute these values:
(-1.6 × 10^-19 C)V = (1/2)(9.11 × 10^-31 kg)(3.70 × 10^5 m/s)^2.
Simplifying the equation, we can solve for V:
V = [(1/2)(9.11 × 10^-31 kg)(3.70 × 10^5 m/s)^2] / (-1.6 × 10^-19 C).
By plugging in the values and performing the calculations, we find:
V ≈ -214.88 volts.
The potential difference needed to stop the electron is approximately 214.88 volts. Note that the negative sign indicates that the electron is moving opposite to the direction of the electric field, as expected for a negatively charged particle.