Einstein's Photoelectric Equation: Energy Quantum of Radiation

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Work function $W_0=\dfrac{hc}{\lambda_0}$ (using the threshold wavelength). $\;\dfrac{\phi_A}{\phi_B}=\dfrac{hc}{\lambda_1}\times\dfrac{\lambda_2}{hc}=\dfrac{\lambda_2}{\lambda_1}$

2. 2.↑ back to the question

   From the graph, for $P$ the threshold frequency $\nu_{0P}=0.1\times10^{15}$ s$^{-1}$, and for $Q$, $\nu_{0Q}=1\times10^{15}$ s$^{-1}$. Since $\nu_{0Q}>\nu_{0P}$, and threshold wavelength $\lambda_0=c/\nu_0$ is inversely related to threshold frequency, $Q$ has the SMALLER threshold wavelength.

3. 3.↑ back to the question

   The PHOTOELECTRIC EFFECT shows the quantum (particle) nature of electromagnetic radiation.

4. 4.↑ back to the question

   Einstein's equation: $\;\dfrac12mv_m^2=h\nu-\phi_0=\dfrac{hc}\lambda-\phi_0$ $\;v_m^2=\dfrac{2hc}m\cdot\dfrac1\lambda-\dfrac{2\phi_0}m$ This is linear in $1/\lambda$ with slope $\dfrac{2hc}m=\tan\theta$, giving $\;h=\dfrac{m\tan\theta}{2c}$. The line intersects the $1/\lambda$ axis (where $v_m^2=0$) at $1/\lambda_0$, the threshold-wavelength reciprocal, where $\phi_0=\dfrac{hc}{\lambda_0}$. Substituting the value of $h$ found above: $\;\phi_0=\dfrac{m\tan\theta}{2c}\times\dfrac c{\lambda_0}=\dfrac{m\tan\theta}{2\lambda_0}$, where $1/\lambda_0$ is the graph's intercept on the $1/\lambda$ axis.

5. 5.↑ back to the question

   In the photon picture, light of frequency $\nu$ consists of photons each carrying a FIXED quantum of energy $h\nu$. In the photon-electron interaction, an electron absorbs the ENTIRE energy of one photon (not fractions from many photons). Photoemission occurs only if this single-photon energy $h\nu$ exceeds the work function $W_0$ of the metal, i.e. only if $\nu>\nu_0=W_0/h$. Increasing the intensity of light (at a fixed frequency below threshold) only increases the NUMBER of photons (each still of energy $h\nu<W_0$) striking the surface per second — it does not increase the energy carried by any individual photon. Since no single photon then carries enough energy to free an electron, no photoemission occurs no matter how intense the light is. Hence it is the frequency, not the intensity, that determines whether emission occurs.

6. 6.↑ back to the question

   (a) By Einstein's equation, $\;eV_0=\dfrac{hc}\lambda-\phi_0\;\Rightarrow\;V_0=\dfrac{hc}e\cdot\dfrac1\lambda-\dfrac{\phi_0}e$. This is a straight line in $V_0$ vs $1/\lambda$ with SLOPE $=\dfrac{hc}e$. So $\;h=\dfrac{e\times\text{slope}}c$, using the measured slope of the line, $e$ and $c$. (b) Increasing the distance between the light source and metal $A$ only decreases the INTENSITY of the light reaching the surface (by the inverse-square law); the frequency (and hence $1/\lambda$) of the light is unchanged. Since the stopping potential depends only on frequency (not intensity), the stopping potential for electrons emitted from $A$ remains UNCHANGED — only the photoelectric current would decrease.

7. 7.↑ back to the question

   (i) The maximum kinetic energy of photoelectrons is independent of the intensity of incident light but varies linearly with its frequency — explained since $K_{max}=h\nu-W_0$ depends only on $\nu$, while intensity only changes the number of photons (hence photocurrent). (ii) There exists a threshold frequency $\nu_0=W_0/h$ below which no photoemission occurs, regardless of intensity, and emission is instantaneous once $\nu>\nu_0$ — since each electron absorbs one photon's full energy at once rather than accumulating energy gradually.

8. 8.↑ back to the question

   Energy of incident photon: $\;E=\dfrac{hc}\lambda=\dfrac{1242\text{ eV nm}}{412.5\text{ nm}}\approx3.0\text{ eV}$ Photoemission occurs only when $E>W_0$. Since $E=3$ eV exceeds the work functions of Na (1.92 eV) and K (2.15 eV), but is LESS than those of Ca (3.20 eV) and Mo (4.17 eV): metals Na and K will show photoelectric emission, while Ca and Mo will not.

9. 9.↑ back to the question

   By Einstein's equation, $\;eV=h\nu-\phi\;\Rightarrow\;e\Delta V=h\Delta\nu$ (on taking the difference between two points on the line). Using the two points $(5\times10^{14}\text{ Hz},0)$ and $(8\times10^{14}\text{ Hz},1.23\text{ V})$ read off the graph: $\;h=\dfrac{e\Delta V}{\Delta\nu}=\dfrac{1.6\times10^{-19}\times(1.23-0)}{(8-5)\times10^{14}}=\dfrac{1.6\times10^{-19}\times1.23}{3\times10^{14}}\approx6.56\times10^{-34}\text{ J s}$

10. 10.↑ back to the question

    (a) (i) Metals such as zinc or cadmium require UV light (higher photon energy, since they have relatively high work functions) to show photoelectric emission. (ii) Metals such as lithium or sodium (lower work functions) show photoelectric emission even with visible light. (b) $K_{max}=6.06\times10^{-19}\text{ J}=\dfrac{6.06\times10^{-19}}{1.6\times10^{-19}}=3.7875\text{ eV}\approx3.78\text{ eV}$ By Einstein's equation: $\;h\nu=K_{max}+W_0=3.78+4.50=8.28\text{ eV}=8.28\times1.6\times10^{-19}\text{ J}=1.325\times10^{-18}\text{ J}$ $\;\nu=\dfrac{1.325\times10^{-18}}{6.63\times10^{-34}}\approx2.0\times10^{15}\text{ Hz}$

11. 11.↑ back to the question

    (a) $\;E_k=h\nu-\phi_0$. This is a straight line of slope $h$ in the $E_k$ vs $\nu$ graph, cutting the $\nu$-axis at the threshold frequency $\nu_0=\phi_0/h$; for $\nu<\nu_0$, $E_k$ would be negative, i.e. no emission occurs — the graph is drawn only for $\nu\ge\nu_0$, rising linearly from the point $(\nu_0,0)$. (b) (I) Yellow light has a HIGHER frequency (shorter wavelength) than red light, hence a higher photon energy $h\nu$. Since $E_k=h\nu-\phi_0$, photoelectrons emitted by yellow light will have MORE kinetic energy $E_k$ than those emitted by red light. (II) The number of photoelectrons emitted per second depends only on the intensity of light (i.e. the number of photons per second), not on frequency. Since yellow and red light here have the SAME intensity, the SAME number of electrons will be emitted in both cases (provided both frequencies exceed threshold).

12. 12.↑ back to the question

    $\;\dfrac{hc}\lambda=K_{max}+\phi$ For case I: $\;\dfrac{hc}{\lambda_1}=K+\dfrac{hc}{\lambda_T}\;\Rightarrow\;hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_T}\right)=K\quad\dots(i)$ For case II: $\;\dfrac{hc}{\lambda_2}=2K+\dfrac{hc}{\lambda_T}\;\Rightarrow\;hc\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_T}\right)=2K\quad\dots(ii)$ Substituting (i) into (ii): $\;hc\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_T}\right)=2hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_T}\right)$ $\;\dfrac1{\lambda_2}-\dfrac1{\lambda_T}=\dfrac2{\lambda_1}-\dfrac2{\lambda_T}\;\Rightarrow\;\dfrac1{\lambda_T}=\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\;\Rightarrow\;\lambda_T=\dfrac{\lambda_1\lambda_2}{2\lambda_2-\lambda_1}$ Work function, $\;W_0=\dfrac{hc}{\lambda_T}=\dfrac{hc(2\lambda_2-\lambda_1)}{\lambda_1\lambda_2}$

13. 13.↑ back to the question

    (i) According to Einstein's photoelectric equation, when light falls on a metal surface, incident photons are absorbed completely by valence electrons of atoms on its surface. The energy $h\nu$ of each photon is partially utilized by an electron to become free (overcome its "work function" $W_0$), and the rest of the absorbed energy provides the maximum kinetic energy to the photoelectron during emission, i.e. $\;h\nu=\dfrac12mv_{max}^2+W_0$. The minimum frequency of incident radiation below which the photoelectric emission stops (i.e. kinetic energy of photoelectron is zero) is called threshold frequency, $\nu_0=W_0/h$. (ii) For $\lambda=3300$ Å: $\;E=\dfrac{hc}\lambda=\dfrac{1.989\times10^{-25}}{3300\times10^{-10}}\text{ J}=\dfrac{6.03\times10^{-19}}{1.6\times10^{-19}}\approx3.77\text{ eV}$ Since $E(3.77\text{ eV})<W_{0,Mo}(4.17\text{ eV})$ and $E<W_{0,Ni}(5.15\text{ eV})$: Mo and Ni will NOT cause photoelectric emission (Na and K, with lower work functions, will). Bringing the laser source nearer (to 50 cm) only increases the INTENSITY of light reaching the metal, not the photon energy or frequency. Since photoemission depends on frequency, not intensity, this will NOT cause Mo or Ni to emit photoelectrons.

14. 14.↑ back to the question

    (i) Two important features of Einstein's photoelectric equation, $K_{max}=h\nu-\phi_0$: (a) below the threshold frequency $\nu_0$ corresponding to work function $\phi_0$, no emission of photoelectrons takes place, regardless of intensity; (b) the maximum kinetic energy of the emitted photoelectron depends only on the frequency of the incident radiation, not on its intensity. (ii) Since there is no photoemission from $P$, the incident radiation's frequency ($10^{15}$ Hz) is less than the threshold frequency of $P$. Photoemission occurs from $Q$ but with zero kinetic energy, i.e. the incident radiation's energy is JUST sufficient to free the electron from $Q$, meaning the incident frequency equals the threshold frequency of $Q$: $\;\nu_{0Q}=10^{15}$ Hz. Work function of $Q$: $\;W_{0Q}=h\nu_{0Q}=6.6\times10^{-34}\times10^{15}=6.6\times10^{-19}\text{ J}=\dfrac{6.6\times10^{-19}}{1.6\times10^{-19}}\approx4.125\text{ eV}$

15. 15.↑ back to the question

    For a given frequency of incident radiation, the stopping potential is independent of its intensity, i.e. the maximum kinetic energy of photoelectrons depends on the light source and the emitter-plate material (through the frequency and the work function), but is independent of the intensity of radiation — since $K_{max}=eV_0=h\nu-W_0$ depends only on $\nu$ and $W_0$, both of which are unaffected by intensity (intensity only changes the number of photons, and hence the number of photoelectrons, per second).

16. 16.↑ back to the question

    According to the photon picture of light, in the photoelectric effect an electron absorbs a quantum of energy ($h\nu$) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (the work function $\phi_0$), the electron is emitted with some kinetic energy, the maximum value of which is given by $\;K_{max}=h\nu-\phi_0$ This is known as Einstein's photoelectric equation. Here $\phi_0$ is the work function of the metal, which is the minimum energy needed by a surface electron to come out. Two features of the photoelectric effect which cannot be explained by wave theory: (i) The instantaneous emission of photoelectrons (no time lag). (ii) The existence of a threshold frequency for a metal surface, below which no emission occurs regardless of intensity.

17. 17.↑ back to the question

    (i) From the graph, metal $Q$'s line crosses the frequency axis at a higher frequency than metal $P$'s line, so $Q$ has the HIGHER threshold frequency, $\nu_{0Q}\approx3\times10^{14}$ Hz (compared to $\nu_{0P}\approx2\times10^{14}$ Hz). (ii) Work function of $Q$ (the metal with the greater threshold frequency, hence greater work function): $\;W_{0Q}=h\nu_{0Q}=6.63\times10^{-34}\times3\times10^{14}\approx1.99\times10^{-19}\text{ J}\approx1.24\text{ eV}$ (iii) $\;K_{max}=h(\nu-\nu_{0Q})=6.63\times10^{-34}\times(8-3)\times10^{14}=3.315\times10^{-19}\text{ J}\approx2.07\text{ eV}$

18. 18.↑ back to the question

    Both graphs are parallel straight lines (same slope $h/e$) of stopping potential $V_0$ versus frequency $\nu$; the line for metal $A$ crosses the frequency axis at the higher threshold frequency $\nu_A$, while the line for $B$ crosses at the lower $\nu_B$, with $A$'s line lying to the right of (below/parallel-shifted from) $B$'s line. (i) For the SAME frequency of incident radiation (above both thresholds), metal $B$ (with the smaller threshold frequency, hence smaller work function) will have a MORE stopping potential than $A$, since $V_0=\dfrac he\nu-\dfrac{W_0}e$ and $W_0$ is smaller for $B$. (ii) No. The slope of the graph is $h/e$, which is a ratio of two universal constants — it does NOT depend on the nature of the material used (only the intercept, i.e. the work function, does).

19. 19.↑ back to the question

    Cut-off (threshold) frequency is the minimum frequency of light below which the photoelectric emission stops completely, however large the intensity of light may be. $\;h(2f-f)=\dfrac12mv_1^2\;\Rightarrow\;hf=\dfrac12mv_1^2\quad\dots(i)$ $\;h(5f-f)=\dfrac12mv_2^2\;\Rightarrow\;4hf=\dfrac12mv_2^2\quad\dots(ii)$ Dividing (i) by (ii): $\;\dfrac14=\dfrac{v_1^2}{v_2^2}\;\Rightarrow\;\dfrac{v_1}{v_2}=\sqrt{\dfrac14}=\dfrac12$ $\;v_1:v_2=1:2$

20. 20.↑ back to the question

    Einstein's photoelectric equation: $\;h\nu=\dfrac12mv_{max}^2+W_0$, where $\nu$ is the frequency of incident radiation, $\dfrac12mv_{max}^2$ is the maximum kinetic energy of an emitted electron, and $W_0$ is the work function of the target metal. Three salient features explained by this equation (which wave theory could not explain): (i) Below the threshold frequency $\nu_0$ corresponding to $W_0$, no emission of photoelectrons takes place — since $h\nu-W_0$ would then be negative, which is not physically allowed. (ii) As energy of a photon depends only on the frequency of light, the maximum kinetic energy with which a photoelectron is emitted depends only on the energy of the photon, i.e. on the frequency of incident radiation — not on its intensity. (iii) For a given frequency of incident radiation, intensity of light depends on the number of photons falling per unit area per unit time, and since ONE photon liberates exactly one photoelectron, the number of photoelectrons emitted depends only on the intensity of incident light, explaining why the photoelectric current is proportional to intensity while $K_{max}$ is not.

21. 21.↑ back to the question

    Einstein's photoelectric equation: $\;K_{max}=h\nu-\phi_0=\dfrac{hc}\lambda-\phi_0$ Features explained: existence of threshold frequency (no emission below it, regardless of intensity), instantaneous emission, and $K_{max}$ depending only on frequency (not intensity) while photocurrent depends only on intensity (not frequency, above threshold). Same numerical derivation as Q40/Q49 pattern (Q40 above): for wavelength $\lambda_1$, $\;\dfrac{hc}{\lambda_1}=K+\dfrac{hc}{\lambda_0}$; for $\lambda_2$ (giving $2K$), $\;\dfrac{hc}{\lambda_2}=2K+\dfrac{hc}{\lambda_0}$. Eliminating $K$: $\;2\left(\dfrac{hc}{\lambda_1}-\dfrac{hc}{\lambda_0}\right)=\dfrac{hc}{\lambda_2}-\dfrac{hc}{\lambda_0}$ $\;\phi_0=\dfrac{2hc}{\lambda_1}-\dfrac{hc}{\lambda_2}=hc\left(\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\right)$ Also $\;\phi_0=\dfrac{hc}{\lambda_0}\;\therefore\;\dfrac{hc}{\lambda_0}=hc\left(\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\right)\;\Rightarrow\;\dfrac1{\lambda_0}=\dfrac{2\lambda_2-\lambda_1}{\lambda_1\lambda_2}\;\Rightarrow\;\lambda_0=\dfrac{\lambda_1\lambda_2}{2\lambda_2-\lambda_1}$ Work function: $\;W_0=\dfrac{hc(2\lambda_2-\lambda_1)}{\lambda_1\lambda_2}$

22. 22.↑ back to the question

    (i) By Einstein's equation, $\;eV_0=h\nu-\phi_0\;\Rightarrow\;V_0=\dfrac he\nu-\dfrac{\phi_0}e$. The slope of the line is $\;\dfrac{\Delta V_0}{\Delta\nu}=\dfrac he$, a ratio of two universal constants ($h$ and $e$). It is therefore a CONSTANT quantity, independent of the nature of the material — hence both lines have the same slope. (ii) At the same frequency $\nu$ (from the graph, $V_1>V_2$, i.e. $M_1>M_2$ in stopping potential), the maximum kinetic energy is $\;K_{max}=eV_0=h\nu-h\nu_0$. Since $M_1$ has a smaller threshold frequency (hence smaller work function), $M_1$'s emitted electrons have GREATER kinetic energy for a given incident frequency.

23. 23.Answer: (B)↑ back to the question

    $K=h\nu-W_0\;\Rightarrow\;h\nu=K+W_0\;\;\dots(ii)$ For doubled frequency: $\;K'=h(2\nu)-W_0=2h\nu-W_0$ Substituting $h\nu=K+W_0$: $\;K'=2(K+W_0)-W_0=2K+W_0$ Since $W_0>0$, $K'=2K+W_0$ is MORE THAN $2K$. Option (b).

24. 24.Answer: (C)↑ back to the question

    $E=h\nu=\dfrac{hc}\lambda\;\Rightarrow\;\lambda=\dfrac{hc}E$. Option (c).

25. 25.Answer: (C)↑ back to the question

    $(KE)_{max}=h\nu-\phi$ $\;\dfrac{(KE)_1}{(KE)_2}=\dfrac{h\nu_1-\phi}{h\nu_2-\phi}=\dfrac{1-0.5}{2-0.5}=\dfrac{0.5}{1.5}=\dfrac13$ Ratio is $1:3$. Option (c).

26. 26.Answer: (B)↑ back to the question

    $E=h\nu=h\nu_{0A}+K_A=h\nu_{0B}+K_B$ $\;h\nu=\dfrac{h\nu}2+K_A=\dfrac{h\nu}3+K_B\;\Rightarrow\;K_A=\dfrac{h\nu}2,\;K_B=\dfrac{2h\nu}3$ $\;\dfrac{K_A}{K_B}=\dfrac{h\nu}2\times\dfrac3{2h\nu}=\dfrac34$. Option (b).