Q1 The threshold wavelength is the minimum wavelength that releases electrons from the metal. \[KE = 0 = h\nu - \Phi \tag{1} \] Using \( \lambda\nu = c \) \[ \Phi = \dfrac{hc}{\lambda} \tag{2} \] \[ \lambda = \dfrac{hc}{\Phi} \tag{3} \] Insert the numbers and the answer is: \[\lambda = \dfrac{(6.626 \times 10^{-34} \; J s) (3 \times 10^8 \; m/s)}{8.01 \times 10^{-19} \; J} \tag{4} \] \[\lambda = 2.482 \times 10^{-7} \; m = 248 \; nm \tag{5}\] Does this make sense? In the problem \( \lambda = 162 \; nm \) ejected electrons from the surface with lots of KE so the threshold wavelength should have lower energy than \(162 \; nm \). Lower energy means larger wavelength and yes our answer is larger than \(162 \; nm \). The work function was already given in the problem you just have to do convert the units. \[\Phi = 8.01 \times 10^{-19} \; J s = 8.01 \times 10^{-19} \; J s \times \dfrac { 1\;eV}{ 1.602 \times 10^{-19} \; J s} = 4.99 \; eV s \tag{6} \] Q2 What is orthonormal? Wavefunctions should be normalized so that the total area is equal to 1 \[ 1 = \int_{-\infty}^{\infty} \psi_n^* \psi_n d\tau \tag{1} \] And wavefunctions should be orthogonal so that when different wavefunctions are compared the area is zero. Can be referred to as being perpendicular because the projection of one wavefunction onto another is zero. \[0 = \int_{-\infty}^{\infty} \psi_n^* \psi_m d\tau \tag{2} \] For the particle in a box normalization has been done. \[ 1 = \int_{0}^{L} \sqrt{\dfrac{2}{L}} \sin(\dfrac{n \pi x}{L}) \sqrt{\dfrac{2}{L}} \sin(\dfrac{n \pi x}{L}) dx \tag{3} \] \[ 1 = \dfrac{2}{L} \int_{0}^{L} \sin^2(\dfrac{n \pi x}{L}) dx \tag{4} \] That integral requires trig substitution and u substitution so we gave you the indefinite version on the front formula sheet. \[1 = \dfrac{2}{L} \Bigg|_0^L \dfrac{x}{2} - \dfrac{\sin(\dfrac{n \pi x}{L})}{4\dfrac{n \pi }{L}} \tag {5} \] \[1 = \dfrac{2}{L} \dfrac{L}{2} \tag {6} \] Yes the functions are normalized. Are they orthogonal? \[ 0 = \int_{0}^{L} \sqrt{\dfrac{2}{L}} \sin(\dfrac{n \pi x}{L}) \sqrt{\dfrac{2}{L}} \sin(\dfrac{m \pi x}{L}) dx \tag{7} \] \[ 0 = \int_{0}^{L} \sin(\dfrac{n \pi x}{L}) \sin(\dfrac{m \pi x}{L}) dx \tag{8} \] Integral also requires trig substitution so we gave it to you. \[ 0 = \Bigg|_0^L \dfrac{\dfrac{m \pi x}{L} \sin(\dfrac{n \pi x}{L}) \cos(\dfrac{m \pi x}{L}) - \dfrac{n \pi x}{L} \sin(\dfrac{m \pi x}{L}) \cos(\dfrac{n \pi x}{L})}{\dfrac{n^2 \pi^2 }{L^2} - \dfrac{m^2 \pi^2 }{L^2}} \tag{9} \] The expression \(sin(\dfrac{q \pi x}{L})\) is 0 at 0 and 0 at L so the whole expression goes to 0 and the wavefunctions are orthonormal. Q3 The intensity of the light is a measure of the total number of photons. The Sun's blackbody spectrum is not normalized so we have: \[0.4 = \dfrac{\int_{400}^{700} \rho(\lambda, T) d\lambda}{\int_{0}^{\infty} \rho(\lambda, T) d\lambda} \tag{1} \] Where \( T\) equals the surface temperature of the sun about \( 5000 K \) The Planck distribution \( \rho (\lambda, T) \) is \[ \rho(\lambda,T) = \dfrac {8\pi hc}{\lambda^5} \cdot \dfrac {1}{e^{\frac {hc}{\lambda k_B T}}-1} \tag{2} \] Q4 We have the force constant using the simple equation \[\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \tag{1}\] \[\mu = \dfrac{79 \times 79}{79 + 79} \; amu = \dfrac{79}{2} = 39.5 \;amu = 6.559 \times 10^{-26} \; kg \tag{2} \] \[\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{240 \; N/m}{6.559 \times 10^{-26} \; kg}} \tag{3} \] \[\nu = 9.63 \times 10^{12} Hz = 9.63 THz \tag{4} \] The zero point energy of the harmonic oscillator is when the quantum number \( v = 0 \) \[ E = \dfrac{1}{2} h \nu \tag{5} \] \[E = \dfrac{1}{2} (6.626 \times 10^{-34} \; J s) (9.63 \times 10^{12} Hz) = 3.19 \times 10^{-21} \; J \tag{6} \] Q5 First set the equilibirum of the harmonic oscillator to \( x = 0 \) The lowest quantum state is \( v = 0 \) \[\psi_0 = \Big(\dfrac{\alpha}{\pi}\Big) ^{1/4} e^{- \dfrac{\alpha x^2}{2}} \tag{1} \] This is a Gaussian function which is symmetrical around 0 and there is no preferred direction for the oscillator to vibrate. Therefore we know \( \langle x \rangle = 0\) and \( \langle p \rangle = 0 \) For \( \langle x^2 \rangle \) and \( \langle p^2 \rangle \) INTEGRATE. \[\langle x^2 \rangle = \int_{-\infty}^{\infty} \Big(\dfrac{\alpha}{\pi}\Big) ^{1/4} e^{- \dfrac{\alpha x^2}{2}} \times x^2 \times \Big(\dfrac{\alpha}{\pi}\Big) ^{1/4} e^{- \dfrac{\alpha x^2}{2}} dx \tag{2}\] \[\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} e^{- \alpha x^2} dx \tag{3}\] Because it is symmetrical around 0 \[\langle x^2 \rangle = 2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} \int_{0}^{\infty} x^2 e^{- \alpha x^2} dx \tag{4}\] \[\langle x^2 \rangle = 2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} \times \dfrac{1}{2^2 \alpha} \Big(\dfrac{\pi}{\alpha}\Big) ^{1/2} \tag{5}\] \[\langle x^2 \rangle = \dfrac{1}{2\alpha} \tag{6}\] Find the average of the momentum squared now: \[\langle p^2 \rangle = \int_{-\infty}^{\infty} \Big(\dfrac{\alpha}{\pi}\Big) ^{1/4} e^{- \dfrac{\alpha x^2}{2}} \times -\hbar^2 \dfrac{d^2}{dx^2} \times \Big(\dfrac{\alpha}{\pi}\Big) ^{1/4} e^{- \dfrac{\alpha x^2}{2}} dx \tag{7}\] \[\langle p^2 \rangle = -\hbar^2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} \int_{-\infty}^{\infty} e^{- \dfrac{\alpha x^2}{2}} \times \alpha(\alpha x^2 - 1) e^{- \dfrac{\alpha x^2}{2}} dx \tag{8}\] \[\langle p^2 \rangle = -\hbar^2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} \int_{-\infty}^{\infty} \alpha(\alpha x^2 - 1) e^{- \alpha x^2} dx \tag{9}\] \[\langle p^2 \rangle = -2\hbar^2 \alpha^2 \Big(\dfrac{\alpha}{\pi}\Big) ^{1/2} \int_{0}^{\infty} x^2 e^{- \alpha x^2} dx + 2\hbar^2 \alpha \Big(\dfrac{\alpha}{\pi}\Big)^{1/2} \int_{0}^{\infty} e^{- \alpha x^2} dx \tag{10}\] \[\langle p^2 \rangle = \dfrac{-\hbar^2 \alpha}{2} + {\hbar^2 }{\alpha} \tag{11}\] \[\langle p^2 \rangle = \dfrac{\hbar^2 \alpha}{2} \tag{12}\] We need to have that \(\Delta{q^2} = \langle q^2 \rangle - \langle q \rangle^2 \) to check on the uncertainty \[\Delta{x^2} = \langle x^2 \rangle - \langle x \rangle^2 = \dfrac{1}{2\alpha} - 0 \tag{13} \] \[\Delta{x} = \dfrac{1}{\sqrt{2\alpha}} \tag{14} \] \[\Delta{p^2} = \langle p^2 \rangle - \langle p \rangle^2 = \dfrac{\hbar^2 \alpha}{2} - 0 \tag{15} \] \[\Delta{p} = \dfrac{\hbar \sqrt{\alpha}}{\sqrt{2}} \tag{16} \] Finally we have Heisenberg \[\Delta{p}\Delta{x} = \dfrac{1}{\sqrt{2\alpha}} \dfrac{\hbar \sqrt{\alpha}}{\sqrt{2}} = \dfrac{\hbar}{2} \tag{17}\] And being equal to \( \dfrac{\hbar}{2} \) is perfectly acceptable and the lowest possible value. So the Gaussian functions that make up the lowest harmonic oscillator have the most accurate position and momentum that is possible in the quantum world. Q6 The important part of this question is understanding that the average momentum is not the instant momentum and that in a Particle in a Box since there is no potential energy all the energy is kinetic. \[ \lambda_{Brog} = \dfrac{h}{p} \tag{1}\] \[ E = KE = \dfrac{h^2 n^2}{8mL^2} = \dfrac{1}{2}\dfrac{p^2}{m} \tag{2}\] \[ p = \dfrac{h n}{2L} \tag{3}\] \[ \lambda_{Brog} = \dfrac{h}{\dfrac{h n}{2L} } = \dfrac{2L}{n} \tag{4}\] For \( n = 2\) we have \( \lambda_{Brog} = L \) And if you draw the wavefunction you see that the wavelength is indeed equal to L. If you need math the wavelength is found from the argument of the \(\sin() \) function \[ \psi = \sqrt{\dfrac{2}{L}} \sin(\dfrac{n \pi x}{L}) \tag{5}\] \[\lambda_{Sch} \dfrac{n \pi }{L} = 2\pi \tag{6} \] \[\lambda_{Sch} = \dfrac{2L}{n} \tag{7} \]