Answer:
A) ΔV = 1.237 V
B) K.E = 1.237 eV
Explanation:
B)
The initial kinetic energy of the electron is given by the following formula:
[tex]K.E = \frac{1}{2}mv^2\\\\[/tex]
where,
K.E = Kinetic Energy of electron = ?
m = mass of elctron = 9.1 x 10⁻³¹ kg
v = speed of electron = 660000 m/s
Therefore,
[tex]K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2[/tex]
K.E = 1.98 x 10⁻¹⁹ J
K.E = (1.98 x 10⁻¹⁹ J)([tex]\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}[/tex])
K.E = 1.237 eV
A)
The energy applied by the potential difference must be equal to the kinetic energy of the electron, in order to stop it:
[tex]e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}[/tex]
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}[/tex]
ΔV = 1.237 V
