Experimental Study of Photoelectric Effect

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

   If the frequency of the incident radiation increases, the kinetic energy of the ejected photoelectrons also increases (since $K_{max}=h\nu-\phi_0$). Hence we would need to increase the (negative) retarding potential to stop them, i.e. the stopping potential INCREASES.

2. 2.↑ back to the question

   Curves 1 and 2 have the SAME stopping potential (i.e. the same threshold/x-intercept), and the stopping potential for a given frequency of light depends only on the material (work function), not on the intensity. So curves 1 and 2 are for the same material (differing only in intensity, hence differing saturation currents); curve 3, having a different stopping potential, is for the other material.

3. 3.↑ back to the question

   Yes. Photoelectric current (which is proportional to the number of photoelectrons emitted per second) is directly proportional to the intensity of the incident radiation, for a given frequency of light. This is because greater intensity means a greater number of photons striking the surface per second, and each photon (above the threshold frequency) can liberate one photoelectron, so more electrons are emitted per second.

4. 4.↑ back to the question

   By Einstein's photoelectric equation: $\;eV_{01}=\dfrac{hc}{\lambda_1}-W_0\;$ and $\;eV_{02}=\dfrac{hc}{\lambda_2}-W_0$ Subtracting: $\;e(V_{01}-V_{02})=hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)$ $(6-V_{02})=\dfrac{4.14\times10^{-15}\times3\times10^8}{1.6\times10^{-19}\times1.6\times10^{-19}}\times1.6\times10^{-19}\left(\dfrac1{4\times10^{-7}}-\dfrac1{6\times10^{-7}}\right)$ — using $hc=4.14\times10^{-15}\times3\times10^8=1.242\times10^{-6}$ eV m directly: $\;(6-V_{02})=hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)$ in eV $=1.242\times10^{-6}\left(\dfrac1{4\times10^{-7}}-\dfrac1{6\times10^{-7}}\right)=1.242\times10^{-6}\times\dfrac{2}{12\times10^{-7}}\approx2.07$ eV $\;6-V_{02}\approx2.07\;\Rightarrow\;V_{02}\approx3.93\text{ V}$ So the new stopping potential is approximately $3.93$ V.

5. 5.↑ back to the question

   (i) The number of photons emitted per second $n$ is given by the ratio of the power emitted to the energy of each photon: $\;n=\dfrac Ph\nu=\dfrac{2.0\times10^{-3}}{6.63\times10^{-34}\times6.0\times10^{14}}=\dfrac{2.0\times10^{-3}}{3.978\times10^{-19}}\approx5.03\times10^{15}\text{ photons/s}$ (ii) Photoelectric current is directly proportional to the intensity of incident radiation (for a given frequency): the graph is a straight line through the origin, i.e. photoelectric current increases linearly with intensity of light.

6. 6.↑ back to the question

   (a) By Einstein's photoelectric equation: $\;K.E._{max}=\dfrac{hc}\lambda-W_0$ $\lambda=2000\text{ \text{Å}}=2\times10^{-7}$ m, $W_0=4.2$ eV, $hc=6.63\times10^{-34}\times3\times10^8$ J m. $\;K.E._{max}=\dfrac{6.63\times10^{-34}\times3\times10^8}{2\times10^{-7}\times1.6\times10^{-19}}\text{ eV}-4.2\text{ eV}=(6.2-4.2)\text{ eV}=2.0\text{ eV}$ (b) The maximum kinetic energy (hence the stopping potential) of the emitted electrons does not depend upon the intensity of the incident light — it depends only on the frequency (wavelength) of light and the work function. So if the intensity is doubled at the same wavelength, there is NO change in the stopping potential; only the photoelectric current (number of photoelectrons) increases.

7. 7.↑ back to the question

   Graph: photoelectric current is directly proportional to intensity of light (for a fixed frequency) — a straight line passing through the origin. Energy of the incident photon: $\;E=\dfrac{hc}\lambda=\dfrac{6.6\times10^{-34}\times3\times10^8}{3300\times10^{-10}\times1.6\times10^{-19}}\approx3.75\text{ eV}$ Since $E(3.75\text{ eV})<W_{0,Mo}(4.175\text{ eV})$, molybdenum (Mo) will NOT show photoelectric emission, while sodium (Na, $W_0=2.75$ eV $<3.75$ eV) will. Bringing the laser source nearer only increases the INTENSITY of light (more photons per second), not the energy of each photon. Since the photon energy is still below Mo's work function, no photoemission will occur from Mo even if the source is brought closer; for Na, bringing the source closer will only increase the photoelectric current (more photoelectrons per second), not their kinetic energy.

8. 8.↑ back to the question

   (a) No. Electrons in a metal do not all occupy the same energy level — some are more tightly bound than others (near the surface vs deeper inside), so they lose different amounts of energy escaping. Hence photoelectrons are emitted with a RANGE of kinetic energies, up to a maximum value $K_{max}=h\nu-W_0$. (b) No, the maximum kinetic energy of the emitted electrons depends only on the frequency of the incident radiation and the work function of the metal, NOT on the intensity of the radiation. (c) The number of photoelectrons emitted per second depends only on the INTENSITY of the incident radiation (for a given frequency above threshold) — more intensity means more photons striking the surface per second, hence more photoelectrons.

9. 9.↑ back to the question

   Both graphs are of the same shape — the photoelectric current rises from a common (negative) stopping potential on the $V$-axis and saturates at a constant value for large positive $V$ — since both radiations have the same frequency (hence same stopping potential). The curve reaching a HIGHER saturation current level corresponds to the radiation of HIGHER intensity $I_2$, while the curve with the lower saturation current corresponds to the lower intensity $I_1$ ($I_2>I_1$); both curves cross the voltage axis at the same (negative) stopping potential $-V_0$.

10. 10.↑ back to the question

    The photocurrent rises steeply from a negative (retarding) anode potential $-V_0$ (the stopping potential), where the current is zero, and then increases until it saturates at a constant SATURATION CURRENT for sufficiently large positive anode potential — the standard S-shaped photocurrent-vs-anode-potential curve.

11. 11.↑ back to the question

    (a)(i) Threshold frequency is the minimum frequency of incident radiation below which no photoelectric emission takes place, irrespective of the intensity of light. (ii) Stopping potential is the minimum (negative/retarding) potential applied to the anode w.r.t. the cathode at which the photoelectric current becomes zero. (b) The two graphs, for the same frequency $\nu$ but different intensities $I_1<I_2$, both cross the voltage axis at the SAME stopping potential (since frequency is unchanged), but the curve for $I_2$ saturates at a HIGHER current level than the curve for $I_1$ — otherwise the two curves have identical shape, just scaled vertically.

12. 12.↑ back to the question

    Since $M_2$'s line crosses the frequency axis at a HIGHER threshold frequency than $M_1$'s, $M_2$ has the greater value of threshold frequency, and therefore $M_2$ has the greater value of WORK FUNCTION (since $W_0=h\nu_0$).

13. 13.↑ back to the question

    $\;E=h\nu\;\Rightarrow\;\nu=\dfrac Eh=\dfrac{2\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx4.8\times10^{14}\text{ Hz}$

14. 14.↑ back to the question

    Using $\;eV_0=h\nu-W_0$: (i) As the incident frequency $\nu$ increases, the stopping potential $V_0$ also INCREASES (they are directly, linearly related through the above equation). (ii) There is NO effect on the stopping potential when the intensity of incident radiation changes (at a fixed frequency), since $V_0$ depends only on $\nu$ and $W_0$, not on intensity — only the photoelectric current (saturation value) will change as the intensity changes.

15. 15.↑ back to the question

    Intensity $I=n\,h\nu$ (energy delivered per unit area per unit time), where $n$ is the number of photons incident per unit area per unit time. For the same intensity, $\;n_1h\nu_1=n_2h\nu_2\;\Rightarrow\;\dfrac{n_2}{n_1}=\dfrac{\nu_1}{\nu_2}>1\;\Rightarrow\;n_2>n_1$ (i) Since for the same intensity the number of photons (and hence photoelectrons) per unit area per unit time is LARGER for the smaller frequency, MORE electrons will be emitted in the case of $\nu_2$ (the smaller frequency). (ii) Maximum kinetic energy $K_{max}=h\nu-W_0$ is more for the greater frequency, so $\nu_1$ (the larger frequency) gives photoelectrons of greater maximum kinetic energy.

16. 16.↑ back to the question

    (i) Energy of incident radiation: $\;E=h\nu=6.63\times10^{-34}\times6.4\times10^{14}=4.24\times10^{-19}\text{ J}=\dfrac{4.24\times10^{-19}}{1.6\times10^{-19}}\approx2.65\text{ eV}$ (ii) Maximum kinetic energy: $\;K_{max}=E-W_0=2.65-2.31=0.34\text{ eV}$ (iii) Stopping potential: $\;eV_0=K_{max}\;\Rightarrow\;V_0=0.34\text{ V}$

17. 17.↑ back to the question

    (a) Photoelectric current is proportional to the number of photoelectrons emitted per second, which in turn is proportional to the number of photons incident per second, i.e. to the intensity of light (for a fixed frequency). Hence photocurrent increases with intensity. (b) By Einstein's photoelectric equation, $\;eV_0=h\nu-W_0\;\Rightarrow\;V_0=\dfrac he\nu-\dfrac{W_0}e$. This is linear in $\nu$ with slope $h/e$ — since $h$ and $e$ are universal constants, this slope is the SAME for every photosensitive surface, regardless of the material (only the intercept $-W_0/e$, i.e. the work function, differs between materials).

18. 18.↑ back to the question

    (a) The variable $X$ on the horizontal axis is the COLLECTOR PLATE POTENTIAL. (b) The point $A$ on the horizontal axis represents the STOPPING POTENTIAL (the negative retarding potential at which the photocurrent falls to zero). (c) For three different frequencies (same intensity), the three curves saturate at the SAME photoelectric current (since intensity, hence the number of photons/electrons per second, is unchanged), but cross the potential axis at three DIFFERENT stopping potentials — with $\nu_1>\nu_2>\nu_3$ giving the stopping potentials in the order $-V_{01}$ (most negative) further left, then $-V_{02}$, then $-V_{03}$ (least negative) closer to the origin, since a higher frequency gives more energetic photoelectrons requiring a larger retarding potential to stop. (d) For three different intensities (same frequency), all three curves cross the potential axis at the SAME stopping potential (frequency unchanged), but saturate at three DIFFERENT current levels, with $I_1>I_2>I_3$ giving progressively higher saturation currents.

19. 19.↑ back to the question

    (i) Emission of photoelectrons starts as soon as light falls on the metal surface, i.e. there is no time lag between the incidence of light and the emission of photoelectrons. (ii) The emission of photoelectrons takes place only when the frequency of the incident radiation is above a certain critical value called the threshold frequency $\nu_0$, which is characteristic of the metal emitting the electrons. Above the threshold frequency, the maximum kinetic energy with which photoelectrons are emitted is directly proportional to the frequency of the incident radiation. (iii) The maximum kinetic energy with which a photoelectron is emitted from a metallic surface is independent of the intensity of light and depends only upon its frequency.

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

    By Einstein's photoelectric equation: $\;eV_0=h\nu-\phi\;\Rightarrow\;\phi=h\nu-eV_0$ Here energy of photon $h\nu=3.2$ eV, stopping potential $V_0=1.5$ V, so eV of stopping potential $=1.5$ eV. $\;\phi=3.2-1.5=1.7$ eV. Option (b).