GSO001 Problem 12
Problem Statement
Bohr Atomic Model, Photoelectric Effect, and Circular Motion of an Electron in a Magnetic Field
Problem Statement
In a hydrogen atom, one photon is emitted when an electron transitions from the energy level n=4 to the energy level n=2, following conservation of energy. This photon is directed onto a metal surface with work function W, causing electrons to be emitted by the photoelectric effect.
The electron with the maximum kinetic energy among the emitted photoelectrons is selected and sent into a vacuum space with a uniform magnetic field of magnetic flux density B, in a direction perpendicular to the field. This electron undergoes uniform circular motion due to the Lorentz force.
Find the radius r of the circular orbit.
Let E0 be the absolute value of the energy of the hydrogen atom's ground state (n=1). The energy level for principal quantum number n is given by −n2E0. Let the electron mass be m, the elementary charge be e. The motion may be treated non-relativistically.
Constraints
- E0=2.16×10−18 J
- W=2.05×10−19 J
- m=9.00×10−31 kg
- e=1.60×10−19 C
- B=1.40×10−4 T
Input Format
The orbital radius r of the electron is expressed as A×10−2 m. Express A as an irreducible fraction qp and find the value of p+q.
Solution
This problem combines three major topics: energy levels in the Bohr model, the photoelectric effect, and the motion of a charged particle in a magnetic field.
1. Computing the Photon Energy
The photon energy ΔE emitted in the transition from n=4 to n=2 (Balmer series):
ΔE=E4−E2=(−42E0)−(−22E0)=E0(41−161)=163E0Substituting E0=2.16×10−18 J:
ΔE=163×2.16×10−18=4.05×10−19 J2. Maximum Kinetic Energy of the Photoelectron
By Einstein's photoelectric equation, the maximum kinetic energy Kmax of the emitted photoelectron:
Kmax=ΔE−W=4.05×10−19−2.05×10−19=2.00×10−19 J3. Computing the Orbital Radius
The speed v of the electron with maximum kinetic energy Kmax:
v=m2Kmax⋯(5)When this electron enters the magnetic field B perpendicularly, it undergoes uniform circular motion with the Lorentz force evB as the centripetal force:
mrv2=evB⇒r=eBmv=eB2mKmaxComputing the numerator:
2mKmax=2×(9.00×10−31)×(2.00×10−19)=36.0×10−50 kg2⋅m2/s2 2mKmax=36.0×10−50=6.00×10−25 kg⋅m/sComputing the denominator eB:
eB=(1.60×10−19)×(1.40×10−4)=2.24×10−23 C⋅TComputing r:
r=2.24×10−236.00×10−25=2.246.00×10−2 mReducing the fraction:
2.246.00=224600=112300=56150=2875So r=2875×10−2 m, giving A=2875, p=75, q=28.
The required value is p+q=75+28=103.