Question

solve this numerical of
class12 ​

Answer 1

Answer:

Explanation:Charge on proton 1, q1 = 1.6 ×10 - 19 C

Charge on proton 2, q2 = 1.6 ×10 - 19 C

Charge on electron, q3 = - 1.6 ×10 - 19 C

Distance between protons 1 and 2, d1 = 1.5 ×10 - 10 m

Distance between proton 1 and electron, d2 = 1 ×10 - 10 m

Distance between proton 2 and electron, d3 = 1 × 10 - 10 m

The potential energy at infinity is zero.

Potential energy of the system,

V=q1q2/4piEod1+q2q3/4piEod3+q3q1/4piEod2

we know that, 1/4piEo=9x10^9 Nm^2 C^-2

=> V=9x10^9x10^-19x10^-19/10^-10(-(16)^2x1.5+(1.6)^2-(1.6)^2x1.5

on solving, you'll get,

V=-19.2eV

Therefore, the potential energy of the system is - 19.2 eV.