Question

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?

Answer 1

Answer:

why put whole question. Answer can be b calculate here.. Can be solve on paper with proper calculations

Answer 2

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å.

charge on electron , e = -1.6 × 10^-19 C

charge on proton, p = + 1.6 × 10^-19 C

seperation between electron and proton, r = 0.53 A° = 0.53 × 10^-10 m

now potential energy , U = kq1q2/r

= k(e)(p)/r

= 9 × 10^9 × (-1.6 × 10^-19)(1.6 × 10^-19)/(0.53 × 10^-10) J

= - (9 × 2.56 × 10^-29)/(0.53 × 10^-10)

= - 43.47 × 10^-19 J

we know, 1eV = 1.6 × 10^-19 J

so, U = -43.47 × 10^-19/(1.6 × 10^-19) eV

= -27.2 eV

(b) kinetic energy = half of magnitude of potential energy

= |U|/2 = |27.2eV|/2 = 13.6 eV

hence, kinetic energy = 13.eV

so, total energy = K.E + U

= 13.6 eV - 27.2 eV = -13.6eV

now minimum energy required to remove an electron , E = ∆U

= 0 -(-13.6eV) = +13.6 eV

(c) potential energy at 1.06 × 10^-10cn seperation.

U' = (9 × 10^9)(-1.6 × 10^-19)(+1.6 × 10^-19)/(1.06 × 10^-10)

= -21.74 × 10^-19 J

= -21.74 × 10^-19/(1.6 × 10^-19) eV

= -13.585 eV

for (a) : talking -13.585 eV as zero of P.E , then P.E of system = -27.2 eV -(-13.585) = -13.585 eV ≈ -13.6 eV

for (b) ; ∆U = -13.6eV - (-13.6eV) = 0

so, workdone , W = 0