The Workbook

1. \(E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{620 \text{ nm}} = 2\) eV.

   In joule: \(E = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\) J.

2. \(p = \dfrac{h}{\lambda} = \dfrac{6.6 \times 10^{-34}}{620 \times 10^{-9}} \approx 1.07 \times 10^{-27}\) kg m/s.

   Check: \(p = E/c = 3.2 \times 10^{-19} / 3 \times 10^{8} \approx 1.07 \times 10^{-27}\) kg m/s.

\(E = h\nu\), so \(\nu = \dfrac{E}{h} = \dfrac{3.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 8 \times 10^{14}\) Hz.

\(\lambda = \dfrac{c}{\nu} = \dfrac{3 \times 10^{8}}{8 \times 10^{14}} = 3.75 \times 10^{-7}\) m \(= 375\) nm.

Check: \(\lambda = 1240/3.3 \approx 375\) nm.

1. Energy of one photon: \(E = 1240/620 = 2\) eV \(= 3.2 \times 10^{-19}\) J.

   Number per second \(n = \dfrac{P}{E} = \dfrac{3.2 \times 10^{-3}}{3.2 \times 10^{-19}} = 10^{16}\).

2. At \(1240\) nm each photon carries \(1\) eV, half as much. The same power needs twice as many photons: \(n = 2 \times 10^{16}\) per second.

1. \(E = 2\) eV \(= 3.2 \times 10^{-19}\) J per photon. \(n = \dfrac{3.2}{3.2 \times 10^{-19}} = 10^{19}\) per second.

2. At \(10\) m the photons are spread over a sphere of area \(4\pi r^2 = 4\pi \times 100 = 400\pi\) m\(^2\).

   Number crossing \(1\) m\(^2\) per second \(= \dfrac{10^{19}}{400\pi} \approx 8 \times 10^{15}\).

1. Each photon absorbed gives its momentum \(E/c\) to the surface. The pressure is \(p = \dfrac{I}{c} = \dfrac{600}{3 \times 10^{8}} = 2 \times 10^{-6}\) Pa.

   Force \(F = pA = 2 \times 10^{-6} \times 0.5 = 10^{-6}\) N.

2. A reflected photon reverses its momentum, so the surface receives \(2E/c\) from each photon. The pressure is \(\dfrac{2I}{c} = 4 \times 10^{-6}\) Pa and the force is \(2 \times 10^{-6}\) N.

1. The work function is the minimum energy needed to remove an electron from the surface of the metal.

2. \(\Phi = h\nu_0\), so \(\nu_0 = \dfrac{3.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 8 \times 10^{14}\) Hz.

   \(\lambda_0 = \dfrac{hc}{\Phi} = \dfrac{1240}{3.3} \approx 375\) nm.

1. \(E = 1240/310 = 4\) eV.

2. \(K_{\max} = E - \Phi = 4 - 2 = 2\) eV.

3. \(eV_s = K_{\max}\), so \(V_s = 2\) V.

1. \(\lambda_0 = \dfrac{1240}{\Phi}\) nm with \(\Phi\) in eV.

   Cs: \(1240/2.1 \approx 590\) nm. Na: \(1240/2.3 \approx 539\) nm. Cu: \(1240/4.0 = 310\) nm.

2. Light of \(500\) nm carries \(1240/500 = 2.48\) eV per photon.

   \(2.48\) eV is more than the work functions of Cs and Na, so they emit. It is less than \(4.0\) eV, so copper does not emit.

Photon energies: \(1240/248 = 5\) eV and \(1240/310 = 4\) eV.

Einstein's equation twice: \(5 - \Phi = 3\) and \(4 - \Phi = 2\). Both give \(\Phi = 2\) eV.

Threshold wavelength: \(\lambda_0 = 1240/2 = 620\) nm.

1. \(E = 1240/310 = 4\) eV. \(K_{\max} = 4 - 2.2 = 1.8\) eV \(= 1.8 \times 1.6 \times 10^{-19} = 2.88 \times 10^{-19}\) J.

2. \(K_{\max} = \dfrac{1}{2}mv^2\), so \(v^2 = \dfrac{2 \times 2.88 \times 10^{-19}}{9 \times 10^{-31}} = 6.4 \times 10^{11}\).

   \(v = 8 \times 10^{5}\) m/s.

1. Doubling the intensity doubles the number of photons per second. Each photon releases at most one electron, so the number of photoelectrons per second doubles.

2. \(K_{\max} = h\nu - \Phi\) depends only on the frequency and the metal. \(K_{\max}\) does not change.

3. \(V_s = K_{\max}/e\), so the stopping potential does not change.

1. \(310\) nm: \(E = 4\) eV, \(K_1 = 4 - 2 = 2\) eV. \(155\) nm: \(E = 8\) eV, \(K_2 = 8 - 2 = 6\) eV.

   \(K_1 : K_2 = 2 : 6 = 1 : 3\).

2. Halving the wavelength doubles the photon energy, not the kinetic energy. The work function is subtracted first, so \(K_{\max}\) grows by more than a factor of two.

1. The line meets the \(\nu\) axis at \(8 \times 10^{14}\) Hz. Below this frequency there are no photoelectrons, so \(\nu_0 = 8 \times 10^{14}\) Hz.

2. \(V_s = \dfrac{h}{e}(\nu - \nu_0)\), so the slope is \(\dfrac{h}{e}\).

   Slope \(= \dfrac{3.3}{(16 - 8) \times 10^{14}} = 4.125 \times 10^{-15}\) V s.

   \(h = e \times \text{slope} = 1.6 \times 10^{-19} \times 4.125 \times 10^{-15} = 6.6 \times 10^{-34}\) J s.

3. \(\Phi = h\nu_0 = 6.6 \times 10^{-34} \times 8 \times 10^{14} = 5.28 \times 10^{-19}\) J \(= 3.3\) eV.

1. The current becomes zero at \(V = -2\) V for both curves, so \(V_s = 2\) V and \(K_{\max} = 2\) eV.

2. A has the larger saturation current. A higher intensity means more photons per second, so more photoelectrons per second. Curve A is for the higher intensity.

3. \(E = 1240/310 = 4\) eV. \(\Phi = E - K_{\max} = 4 - 2 = 2\) eV.

1. B has the larger stopping potential (\(3\) V against \(2\) V), so its photoelectrons have more energy. A higher \(K_{\max}\) means a higher frequency. Beam B has the higher frequency.

2. For A: \(E = 1240/310 = 4\) eV and \(K_{\max} = 2\) eV. \(\Phi = 4 - 2 = 2\) eV.

3. For B: \(E = \Phi + K_{\max} = 2 + 3 = 5\) eV. \(\lambda = 1240/5 = 248\) nm.

4. The saturation current counts the photoelectrons per second. Both beams deliver the same number of photons per second, so they release the same number of electrons per second.

1. B needs a higher frequency (\(6 \times 10^{14}\) Hz) before any electron comes out. \(\Phi = h\nu_0\), so B has the larger work function.

2. \(\Phi_A = h\nu_A = 6.6 \times 10^{-34} \times 4 \times 10^{14} = 2.64 \times 10^{-19}\) J.

   \(\Phi_B = 6.6 \times 10^{-34} \times 6 \times 10^{14} = 3.96 \times 10^{-19}\) J.

   \(\Phi_A : \Phi_B = 4 : 6 = 2 : 3\).

3. \(K_{\max} = h\nu - \Phi\). The slope is \(h\) for every metal; only the intercept \(\Phi\) changes. So the lines are parallel.

1. The collector A must be at the lower potential. Then A repels the electrons, and only electrons with enough kinetic energy reach it. The potential that stops even the fastest electrons is the stopping potential.

2. When A is positive enough, every photoelectron emitted by C is collected. The current cannot rise above the number of electrons emitted per second. This is the saturation current.

3. At twice the distance the intensity falls to one quarter, so the saturation current falls to one quarter. The frequency is unchanged, so \(K_{\max}\) and the stopping potential are unchanged.

1. Electrons per second \(= 10^{16}/100 = 10^{14}\).

   \(I = ne = 10^{14} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-5}\) A \(= 16\ \mu\)A.

2. Twice the intensity means \(2 \times 10^{16}\) photons per second and \(2 \times 10^{14}\) electrons per second. \(I = 32\ \mu\)A.

1. An electron takes energy from one photon at a time. A red photon carries less than the work function, so no red photon can free an electron, however many there are. One violet photon carries more than the work function, so it frees an electron.

2. The energy arrives in one packet, not slowly over the whole surface. The first photons that arrive can free electrons at once.

\(K_{\max} = h\nu - \Phi\) is a straight line of slope \(h\).

It meets the \(\nu\) axis where \(K_{\max} = 0\): \(\nu_0 = \dfrac{\Phi}{h} = \dfrac{3.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 8 \times 10^{14}\) Hz.

Its extension meets the \(K_{\max}\) axis at \(\nu = 0\): \(K_{\max} = -\Phi = -3.3\) eV. The dashed part below the axis is not physical; no electrons come out there.

1. \(eV_s = K_{\max} = h\nu - \Phi\). Divide by \(e\): \(V_s = \dfrac{h}{e}\nu - \dfrac{\Phi}{e}\).

2. The slope is \(h/e\), a ratio of two universal constants. The metal only changes the intercept \(\Phi/e\).

1. \(E = 1240/200 = 6.2\) eV. \(K_{\max} = 6.2 - 4 = 2.2\) eV, so \(V_s = 2.2\) V.

2. \(E = 1240/400 = 3.1\) eV, which is less than \(\Phi = 4\) eV. No photoelectrons are emitted, so there is no current and no stopping potential.

1. Violet: \(E = 1240/400 = 3.1\) eV.

   Radio: \(\lambda = 1\) m \(= 10^{9}\) nm, so \(E = 1240/10^{9} = 1.24 \times 10^{-6}\) eV.

2. Number \(= \dfrac{3.1}{1.24 \times 10^{-6}} = 2.5 \times 10^{6}\). One violet photon equals \(2.5\) million radio photons.

1. Photon energies: A gives \(2\) eV, B gives \(4\) eV. For the same power, \(n \propto 1/E\).

   \(n_A : n_B = 4 : 2 = 2 : 1\).

2. Only B has photons with energy above \(3\) eV. B causes emission with \(K_{\max} = 4 - 3 = 1\) eV. A gives no emission though it has more photons.