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CHEM-E4110 Quantum mechanics and Spectroscopy Exercise I, Solutions Exercise 1 – The lifespan of a classical atom This exercise can be done using pen and paper. Lets assume that an electron is a point-wise particle, orbiting the nucleus in a spherical trajectory with radius a0 = 0.529 Å. In a spherical orbit the electron 2 experiences only the centripetal acceleration, ~a = ~vr . A charged particle in an accelerated motion emits energy according to the Larmor equation: e2 |~a|2 dE =− dt 6π0 c3 where ~a is the acceleration of the particle. How long does it take for a classical, point-wise electron to collapse into the nucleus? Hint: The acceleration is caused by the electrostatic interaction between the electron and the nucleus. Write the total energy as the sum of kinetic and potential energy: e2 1 2 E = m|v| − 2 4π0 r Find a way to express the total energy as a function of the radius r. After this you can use the Larmor equation and the chain rule df (x) df (x) dx = dt dx dt and integrate the radius from a0 → 0 and time t : 0 → tend . Solution: In the Bohr model for hydrogen atom, an electron is considered to be a classical particle that can orbit the nucleus only in some allowed trajectories. The term classical means that it has well-defined position and momentum at all times (this is not case in quantum mechanics). The total energy of 1 the electron depends on the potential and kinetic energy. Potential energy is determined by the Coulomb potential (electro-static interaction between the electron and the nucleus). Kinetic energy depends on the velocity of the electron. e2 1 E = m|v|2 − 2 4π0 r The kinetic energy can be expressed using the centripetal acceleration: |F~ | e2 v2 = = r m 4π0 mr2 e2 ⇒ v2 = 4π0 mr |~a| = where the force F~ can be obtained by differentiating the potential with respect to r: dV (r) d e2 = dr dr 4π0 r e2 = 4π0 r2 Now the total energy can be written as a function of r: |F~ | = 1 e2 e2 E= m − 2 4π0 mr 4π0 r e2 =− 8π0 r Using the chain rule the change in total energy can be expressed as a time derivative: dE dE dr e2 dr e2 |~a|2 = = = − dt dr dt 8π0 r2 dt 6π0 c3 e2 3c3 dr , |~ a | = dt = − 4|~a|2 r2 4π0 mr2 12π 2 20 m2 c3 2 dt = − r dr e4 The lifespan of the classical atom can be calculated when the above differential is integrated from t = 0 to tend when the electron collapses into the nucleus. During this time the radius changes from r = a0 , when t = 0 to r = 0 when t = tend . For the final calculation one requires the natural constants: c = 299 792 458 m s-1 is the speed of light, 0 = 8.854 · 10-12 C2 N-1 m-2 is vacuum permittivity, e = 1.602 · 10-19 C is an elementary charge and m = 9.109 · 10-31 kg is the mass of the electron. 2 t 0 12π 2 20 m2 c3 2 r dr e4 aH 0 4π 2 20 m2 c3 a3H ∆t = ≈ 15.6 ps e4 Thus, according to classical physics, the hydrogen atom should collapse in 16 pico seconds. The instability of the classical atom is one of the reasons a more comprehensive model for our physical reality was needed and what eventually lead to the development of quantum mechanics. Z Z dt = − 3