Chemistry 432 Problem Set 1 Spring 2018 Solutions

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1 Chemistry 43 Problem Set 1 Spring 018 Solutions 1. A ball of mass m is tossed into the air at time t = 0 with an initial velocity v 0. The ball experiences a constant acceleration g from the gravitational attraction of the earth (g is a positive number). Take the initial location of the ball to be at z = 0. (a) Show that the velocity of the ball at time t is given by v = v 0 gt dv dt = a = g dv = gdt v = v 0 gt (b) Show that the height of the ball at time t is given by z = v 0 t 1/gt dz dt = v = v 0 gt dz = v 0 dt gtdt z = v 0 t 1 gt + z 0 But z = 0 at t = 0. Then z = v 0 t 1 gt 1

2 (c) Use parts a and b to show that v = v 0 gz. v = (v 0 gt) = v 0 + g t v 0 gt = v 0 g(v 0 t 1 gt ) = v 0 gz (d) Use part c to show that the total energy of the ball is conserved during its motion. 1 mv = 1 mv 0 1 m(gz) 1 mv 0 = 1 mv + mgz 1 mv = kinetic energy mgz = potential energy Then kinetic energy + potential energy is a constant.. Show that x(t) = cos ωt satisfies Newton s second law for the motion of a harmonic oscillator in 1 dimension. Evaluate x(t) and the linear momentum of the oscillator at time t = 0. Show that the sum of the potential and kinetic energies of this oscillator is a constant for all times. mẍ + mω x = 0 ẋ(t) = ω sin ωt ẍ(t) = ω cos ωt mω cos ωt + mω cos ωt = 0 x(t = 0) = 1 p(0) = mẋ(t = 0) = 0 E = 1 mẋ + 1 mω x = 1 m( ω sin ωt) + 1 mω cos ωt which is time independent. = 1 mω (sin ωt + cos ωt) = 1 mω

3 3. Consider a particle of mass m in a one-dimensional harmonic well subject to the potential energy V = (1/)mω x. Show that x(t) = A sin ωt + B cos ωt is a solution to Newton s second law for the particle. Introduce the boundary conditions x(0) = x 0 and p(0) = 1 where x 0 is a constant having units of length, to solve for the initial constants A and B. Then show that the sum of the kinetic and potential energies of the particle is independent of time. F = V x = mω x m d x dt = mω x ẋ = Aω cos ωt Bω sin ωt Substituting into the equation of motion d x dt + ω x = 0 ẍ = Aω sin ωt Bω cos ωt Aω sin ωt Bω cos ωt + ω (A sin ωt + B cos ωt) = 0 x(0) = B = x 0 p(0) = mẋ(0) = maω = 1 A = 1 mω x(t) = 1 mω sin ωt + x 0 cos ωt ẋ(t) = 1 m cos ωt ωx 0 sin ωt E = 1 mẋ + 1 mω x = 1 ( ) 1 m m cos ωt ωx 0 sin ωt + 1 ( ) 1 mω mω sin ωt + x 0 cos ωt = 1 ( cos m ωt + ω x m 0 sin ωt ωx 0 cos ωt sin ωt m + 1 ( 1 mω m ω sin ωt + x 0 cos ωt + x ) 0 mω cos ωt sin ωt = 1 m (sin ωt + cos ωt) + mω x 0 (sin ωt + cos ωt) = 1 m + mω x 0 de dt = 0 ) 3

4 4. z=0 A classical particle of mass m and charge q is confined to move between the parallel plates of a capacitor as represented above. The z direction in Cartesian coordinates is taken to be perpendicular to the plates of the capacitor, and the origin of coordinates in the z direction is taken to be the top plate of the capacitor as shown. Between the capacitor plates the electric field, E is a constant, and the potential energy of the charged particle is given by V (z) = qez. Considering only motion in the z direction, give Newton s second law of motion for the particle. Show that the solution to the differential equation associated with Newton s second law is given by z(t) = qet m + c 1t + c where c 1 and c are constants. Evaluate c 1 and c for the initial (boundary) conditions dz/dt = 0 at t = 0 and z = 0 at t = 0. F = dv dz = qe F = ma = m d z dt qe = m d z dt dz dt = qet m + c 1 d z dt = qe m m d z dt = qe which checks ż(t = 0) = 0 + c 1 = 0 or c 1 = 0 z(0) = c = 0 or c = 0 5. The work function of calcium metal is.4 ev. Calculate the kinetic energy of electrons emitted from calcium when light of wavelength nm shines on the surface. K = hc λ φ 4

5 = ( J s)( m s 1 ( ) ) J (.4 ev) = J m ev 6. Find the de Broglie wavelength for an electron, a proton and a 100. g bowling ball each having. ev of kinetic energy. Find the uncertainty in the position of each particle taking p = p. λ = h p p m = E p = (me)1/ Electron p = (me) 1/ = [( kg)(.ev)( J ev 1 )] 1/ = kg m s 1 Proton λ = x = Js kg m s 1 = m = 8.7Å h p = Js 4π( kg m s 1 ) =.69Å p = [( kg)(.0ev)( J ev 1 )] 1/ = kg m s 1 Bowling Ball λ = Js =.0Å kg m s x = λ 4π =.016Å p = [(.1kg)(.0ev)( J ev 1 )] 1/ = kg m s 1 λ = Js kg m s 1 = m x = λ 4π = m 7. A one-dimensional electron is bound by a box to a region of space having length 1.0 Å. A second electron is bound by a similar box but of length.0 Å. Use the uncertainty principle to estimate the ratio of the zero-point kinetic energies of the two electrons. p = p = h x Then KE 1 = p m = KE KE 1 = h 8m( x 1 ) ( ) ( ) x1 1 = = 1 4 x 5

6 8. Consider the expression for the energy of a harmonic oscillator of mass m and angular frequency ω E = 1 mω x + p m. Assuming the uncertainties in the coordinate x and momentum p to be x and p; i.e. write x x p p and using the minimum uncertainty relation x p = h minimize E with respect to x. Show that the energy evaluated at that value of x is E = hω/. As shown later this semester, the value of the energy at the minimum matches the exact ground state energy of the quantum oscillator. E = 1 mω x + p m at or p = h x E = 1 mω x + h 8mx E x = mω x h 4mx = 0 3 mω x = h 4mx 3 x 4 = h 4m ω ( h x = mω ) 1/ E = 1 h mω mω + h mω 8m h = hω 9. Consider a quantum system with total energy given by the expression E = p m + αx 0 x < with α a constant. Use the uncertainty principle to estimate the energy of the system. x p = h p h x 6

7 and E = h 8mx + αx E x = h 4mx + α 3 = 0 at x = ( h 4mα ) 1/3 ( ) E = h 4mα /3 ( h 8m h + α 4mα ) /3 ( hα(1 + ) = m ) 1/3 10. The expression for the energy of a one-dimensional hydrogen atom composed of a proton at a fixed location in space and a moving electron having mass m, coordinate x and momentum p is given by E = p m K x where K is a collection of constants including the charges and other information unimportant in the current problem. Use the uncertainty principle assuming p p and x x and minimize the energy E to estimate the energy of the one-dimensional hydrogen atom. x p h p h x at E = h 8mx K x E x = h 4mx + K 3 x = 0 x = h 4mK E = h 16m K 8m h 4 4mK h = mk h 11. When light of wavelength 4000 Å shines on the surface of barium metal, electrons of kinetic energy.63 ev are emitted. Calculate the wavelength of light needed so that 7

8 photoelectrons of de Broglie wavelength 10 Å are emitted from the surface of barium. hc λ = K + φ ( Js)( ms 1 ) m φ = J p = h λ d = Js m = kg m s 1 = J + φ K = p m = ( kg m s 1 ) kg Js ms 1 λ = J = ( J J) λ = m = 310Å 1. When light of wavelength 550Å shines on a particular surface of tungsten metal, the ejected electrons are found to have a de Broglie wavelength of m. Calculate the kinetic energies of the ejected electrons if light of frequency s 1 shines on the same surface. p = h = J s λ db m = kg m s 1 KE = p m = ( kg m s 1 ) ()( kg) = J KE + φ = hc λ φ = ( J s)( m s 1 ) J = J m KE = hν φ = ( J s)( s 1 ) J = J 13. The work function of cesium metal is.10 ev. Calculate the de Broglie wavelength of photoelectrons emitted from the surface of Cs when light of wavelength nm interacts with the surface. KE = hc λ φ = ( Js)( ms 1 ).10 ev J m ev = J 8

9 p = KE = p p = m e (KE) m e ( kg)( J) = kg m s 1 λ db = h p = Js kg m s 1 = m 14. When electromagnetic radiation of frequency ν = s 1 shines on the surface of metallic nickel, electrons are emitted with debroglie wavelength 11.Å. When radiation of the same frequency shines on sodium metal, the kinetic energy of the emitted electrons is 3.93 ev. Let ν min be the minimum frequency of light required such that electrons are emitted from the surface of metallic nickel. Calculate the kinetic energy of electrons emitted from the surface of sodium metal if radiation of frequency ν min shines on sodium. hν = φ + KE p = h λ = JS m = kg m s 1 φ Ni = hν KE = ( Js)( s) ( kg m s 1 ) ( kg) = J φ Na = ( Js)( s) (3.93 ev)( J ev 1 ) = J hν min = φ Ni KE = hν min φ Na = φ = J J = J 9

CHM320 EXAM #2 USEFUL INFORMATION

CHM320 EXAM #2 USEFUL INFORMATION CHM30 EXAM # USEFUL INFORMATION Constants mass of electron: m e = 9.11 10 31 kg. Rydberg constant: R H = 109737.35 cm 1 =.1798 10 18 J. speed of light: c = 3.00 10 8 m/s Planck constant: 6.66 10 34 Js