Question

Question 2.18: In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-88

Answer 1

(a) in hydrogen , the electron and proton are bound at a distance of about 0.53A° .
so, potential energy between proton and electron is $\bf{U=\frac{q_pq_e}{4\pi\epsilon_0r}}$
here, $q_p$ is charge on proton and $q_e$ is charge on electron. r is the separation between electron and proton.

now, U = 9 × 10^9(1.6 × 10^-19)(-1.6×10^-19)/(0.53× 10^-10) Joule
because 1 Joule = 1.6 × 10^-19 eV
so, U = -9 × 10^9 × (1.6 × 10^-19)²/(1.6 × 10^-19 × 0.53 × 10^-10)
= -9 × 10^9 × 1.6 × 10^-19/(0.53 × 10^-10)
= -27.2 eV

(b) $\bf{E}=+\frac{|U|}{2}$
because kinetic energy is half of the potential energy in case of subatomic particle --> Bohr's theory,
so, Kinetic energy = 27.2/2 = 13.6 eV
so, total energy , E = K.E + P.E
= -27.2 + 13.6 = -13.6 eV
now, Workdone = change in potential energy
= ∆U = potential at infinity - potential at r {of each electron}
[ as you see potential energy of system at r = 0.53A° is -27.2 eV so, potential energy of each electron is -13.6 eV]

so, W = 0 - (-13.6eV) = +13.6 eV

(c) P.E at 1.06 × 10^-10 m separation ,
U' = 9 × 10^9 × (1.6 × 10^-19)(-1.6 × 10^-19)/(1.06 × 10^-10) = -21.74 × 10^-19 J
so, U = -21.74 × 10^-19/1.6 × 10^-19 eV
= -13.585 eV
for (a) , so, potential of the system,∆P.E = potential energy at 0.53 A° - potential energy at 1.06A°
= -27.2 eV -( - 13.585 eV) ≈ -13.6 eV

for (b) , Workdone = potential at 1.06 A° - potential at 0.53A° { of each electron}
= -13.6 - (-13.6 ) = 0
hence, W = 0