Question

An electron with an initial speed of 660,000 m/s is brought to rest by an electric field.

Answer 1

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:

$K.E = \frac{1}{2}mv^2$

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,

$K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2$

K.E = 1.98 x 10⁻¹⁹ J

K.E = (1.98 x 10⁻¹⁹ J)($\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}$)

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:

$e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}$

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

$\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}$

ΔV = 1.237 V