Question

(a) In electron-volts, how much work does an ideal battery with a 17.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.88 × 1018 electrons pass through each second, what is the power of the battery?

Answer 1

(a) 17.0eV

(b) 10.55W

Step-by-step explanation:

(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;

W = q x V --------(i)

Where;

q = charge on the electron = 1e

From the question;

V = 17.0 V

Substitute the values of q and V into equation (i) as follows;

W = 1e x 17.0

W = 17.0eV

Therefore, the work done in electron volts is 17.0

(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;

P = (n q V) / t --------------(ii)

Where;

n = number of electrons = 3.88 x 10¹⁸

t = 1s

q = 1.6 x 10⁻¹⁹C

V = 17.0V

Substitute these values into equation (ii) as follows;

P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1

P = 10.55W

Therefore the power of the battery is 10.55W