The 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

   Circuit diagram of experimental set-up for study of photoelectric effect:

2. 2.↑ back to the question

   Einstein's photoelectric equation: $$K.E._{max}=(h\nu-\phi_0)$$

3. 3.↑ back to the question

   The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, is known as photoelectric effect. i. If increasingly negative potentials were applied to the collector in experiment of photoelectric effect, the photocurrent decreases and for some typical value ($-V_0$), photocurrent becomes zero. This value of $V_0$ is termed as cut-off or stopping potential. ii. The minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface is called the work function of the metal.

4. 4.↑ back to the question

   i. The minimum frequency of incident radiation required to start a photoemission in any photosensitive material is known as threshold frequency. ii. [From 14.2, Q3] The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, is known as photoelectric effect. i. If increasingly negative potentials were applied to the collector in experiment of photoelectric effect, the photocurrent decreases and for some typical value ($-V_0$), photocurrent becomes zero. This value of $V_0$ is termed as cut-off or stopping potential. ii. The minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface is called the work function of the metal..(ii).

5. 5.↑ back to the question

   No. Any light possessing sufficient energy to initiate the photoemission can be used to get photoelectric effect.

6. 6.↑ back to the question

   The wave theory of electromagnetic radiation could not explain following observations of photoelectric effect. i. According to wave theory, the metal surface will require reasonable time to accumulate sufficient energy to knock off electrons. However, the ejection of electrons is an instantaneous process. ii. Since larger incident intensity implies larger energy, the electrons are expected to be emitted with larger kinetic energy. But the observation showed that the maximum kinetic energy did not depend on the incident intensity but depended on the incident frequency. iii. Wave theory expected photoelectrons to be emitted for any frequency. But observations indicated that for a given metal surface, some characteristic cut-off frequency $\nu_0$ existed below which no photoelectrons were emitted however intense the incident radiation was. (Any two points)

7. 7.↑ back to the question

   [From 14.2, Q1] Circuit diagram of experimental set-up for study of photoelectric effect:for Circuit diagram of experimental set-up i. A laboratory experimental set-up for the photoelectric effect consists of an evacuated glass tube with a quartz window. ii. The glass tube contains photosensitive metal plates. One is the emitter E and another plate is the collector C. iii. The emitter and collector are connected to a voltage source whose voltage can be changed and to an ammeter to measure the current in the circuit. iv. A potential difference of V, as measured by the voltmeter, is maintained between the emitter E and collector C. Generally, C (the anode) is at a positive potential with respect to the emitter E (the cathode). This potential difference can be varied and C can even be at negative potential with respect to E. v. When the anode potential (V) is positive, it accelerates the electrons. This potential is called accelerating potential. When the anode potential (V) is negative, it retards the flow of electrons. This potential is known as retarding potential. vi. A source S of monochromatic light of sufficiently high frequency (short wavelength $\leq10^{-7}$ m) is used.

8. 8.↑ back to the question

   [From 14.2, Q2] Einstein's photoelectric equation: $$K.E._{max}=(h\nu-\phi_0)$$for equation. Characteristics of Photoelectric effect: i. Photoemission occurs after the frequency of incident radiation is increased beyond certain minimum value known as threshold frequency $\nu_0$. Threshold frequency is different for different material. ii. Photocurrent is directly proportional to the intensity of incident radiation when frequency and accelerating potential is kept constant. iii. For fixed intensity, the photocurrent increases with increase in voltage until it reaches a saturation value ($I_0$) where it remains constant. iv. There is no time lag between the incidence of light and emission of electrons. The process is instantaneous. v. Maximum kinetic energy of electrons depends on the frequency of incident radiations. vi. Increasingly negative collector potential reduces the photocurrent until it reaches zero at a specific value ($-V_0$), known as the cut-off or stopping potential. This stopping potential is directly proportional to frequency of incident radiation. (Any two characteristics)

9. 9.↑ back to the question

   [From 14.2, Q3] The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, is known as photoelectric effect. i. If increasingly negative potentials were applied to the collector in experiment of photoelectric effect, the photocurrent decreases and for some typical value ($-V_0$), photocurrent becomes zero. This value of $V_0$ is termed as cut-off or stopping potential. ii. The minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface is called the work function of the metal.Experimental set up: [From 14.2, Q1] Circuit diagram of experimental set-up for study of photoelectric effect:, Q. No.7 [(i) – (iv)]

10. 10.↑ back to the question

    Given: $\lambda_0=3800$ Å $=3.8\times10^{-7}$ m, $\lambda=2600$ Å $=2.6\times10^{-7}$ m To find: Maximum kinetic energy (K.E.$_{max}$), Stopping potential ($V_0$) Formulae: i. $K.E._{max}=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right)$ ii. $K.E._{max}=eV_0$ Calculation: From formula (i), $$K.E._{max}=\dfrac{6.63\times10^{-34}\times3\times10^8}{10^{-7}}\left(\dfrac{1}{2.6}-\dfrac{1}{3.8}\right)$$ $$=\dfrac{6.63\times3\times10^{-19}(3.8-2.6)}{2.6\times3.8}=2.42\times10^{-19}\text{ J}=\dfrac{2.42\times10^{-19}}{1.6\times10^{-19}}=1.51\text{ eV}$$ From formula (ii), $eV_0=1.51$ eV $\therefore V_0=1.51$ V Ans: Maximum kinetic energy is 1.51 eV, stopping potential is 1.51 V.

11. 11.↑ back to the question

    Given: $\phi_0=4.2$ eV $=4.2\times1.6\times10^{-19}=6.72\times10^{-19}$ J, $V_0=3$ V, $h=6.63\times10^{-34}$ J-s, $e=1.6\times10^{-19}$ C, $c=3\times10^8$ m/s To find: i. Threshold wavelength ($\lambda_0$) ii. Maximum kinetic energy (K.E.$_{max}$) Formulae: i. $\lambda_0=\dfrac{hc}{\phi_0}$ ii. $V_0=\dfrac{K.E._{max}}{e}$ Calculation: i. From formula (i), $$\lambda_0=\dfrac{hc}{W_0}=\dfrac{6.63\times10^{-34}\times3\times10^8}{6.72\times10^{-19}}=\dfrac{6.63\times3}{6.72}\times10^{-7}$$ $$=\{\text{antilog}[\log(6.63)+\log(3)-\log(6.72)]\}\times10^{-7}$$ $$=\{\text{antilog}[0.8215+0.4771-0.8274]\}\times10^{-7}$$ $$=\{\text{antilog}[0.4712]\}\times10^{-7}$$ $$=2.959\times10^{-7}\text{ m}$$ $\therefore\lambda_0=2959$Å ii. From formula (ii), $K.E._{max}=V_0\times e=3\times1.6\times10^{-19}$ $=4.8\times10^{-19}$ J Ans: i. Threshold wavelength is 2959 Å. ii. Maximum kinetic energy of the emitted electrons is $4.8\times10^{-19}$ J.

12. 12.↑ back to the question

    Given: $E=2.072$ eV $=2.072\times1.6\times10^{-19}$ J, $e=1.6\times10^{-19}$ C, $c=3\times10^8$ m/s, $h=6.63\times10^{-34}$ J-s To find: Wave number ($\bar\nu$) Formula: $\bar\nu=\dfrac{1}{\lambda}=\dfrac{E}{ch}$ Calculation: From formula, $$\bar\nu=\dfrac{2.072\times1.6\times10^{-19}}{3\times10^8\times6.63\times10^{-34}}$$ $$\therefore\bar\nu=\left[\dfrac{2.072\times1.6}{3\times6.63}\right]\times10^7$$ $$=\{\text{antilog}[\log(2.072)+\log(1.6)-\log(3.0)-\log(6.63)]\}\times10^7$$ $$=\{\text{antilog}[0.3164+0.2041-0.4771-0.8215]\}\times10^7$$ $$=\{\text{antilog}(\overline{1}.2219)\}\times10^7$$ $$=0.1666\times10^7$$ $\bar\nu=1.666\times10^6\text{ m}^{-1}$ Ans: The wave number is $1.666\times10^6\text{ m}^{-1}$.

13. 13.↑ back to the question

    Given: $\lambda_0=230$ nm $=230\times10^{-9}$ m, $\lambda=180$ nm $=180\times10^{-9}$ m, $h=6.63\times10^{-34}$ J s, $c=3\times10^8$ m/s To find: K.E. (in joule and eV) Formula: $K.E._{max}=h(\nu-\nu_0)$ Calculation: Using formula, $$K.E._{max}=h\left(\dfrac{c}{\lambda}-\dfrac{c}{\lambda_0}\right)=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right)=hc\left(\dfrac{\lambda_0-\lambda}{\lambda\lambda_0}\right)$$ $$=6.63\times10^{-34}\times3\times10^8\left(\dfrac{1}{180\times10^{-9}}-\dfrac{1}{230\times10^{-9}}\right)$$ $$=\dfrac{6.63\times3\times10^{-26}}{10^{-8}}\left(\dfrac{23-18}{23\times18}\right)=\dfrac{6.63\times3\times5}{23\times18}\times10^{-18}$$ $$=\{\text{antilog}[\log(6.63)+\log(3)+\log(5)-\log(23)-\log(18)]\}\times10^{-18}$$ $$=\{\text{antilog}[0.8215+0.4771+0.6990-1.3617-1.2553]\}\times10^{-18}$$ $$=\{\text{antilog}[1.9976-2.6170]\}\times10^{-18}$$ $$=\{\text{antilog}[\overline{1}.3806]\}\times10^{-18}=0.2402\times10^{-18}$$ $\therefore K.E._{max}=2.402\times10^{-19}$ J $$\therefore K.E._{max}(\text{in eV})=\dfrac{2.402\times10^{-19}}{1.6\times10^{-19}}\text{ eV}$$ $$=\{\text{antilog}[\log(2.402)-\log(1.6)]\}$$ $$=\{\text{antilog}[0.3806-0.2041]\}$$ $$=\{\text{antilog}[0.1765]\}$$ $$=1.5\text{ eV}$$ Ans: The maximum kinetic energy of the ejected electron is $2.402\times10^{-19}$ J or 1.5 eV.

14. 14.↑ back to the question

    Given: $\lambda_0=3800$ Å $=3.8\times10^{-7}$ m, $\lambda=2600$ Å $=2.6\times10^{-7}$ m To find: Maximum kinetic energy (K.E.$_{max}$) Formula: $K.E._{max}=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right)$ Calculation: From formula, $$(K.E.)_{max}=6.63\times10^{-34}\times3\times10^8\times\left[\dfrac{1}{2600\times10^{-10}}-\dfrac{1}{3800\times10^{-10}}\right]$$ $$=6.63\times10^{-34}\times3\times10^8\times\left[\dfrac{(3800-2600)\times10^{-10}}{2600\times3800\times10^{-20}}\right]$$ $$=\dfrac{6.63\times10^{-34}\times3\times10^8\times1200}{2600\times3800\times10^{-10}}=\dfrac{6.63\times36}{26\times38}\times10^{-18}$$ $$=\{\text{antilog}[\log(6.63)+\log(36)-\log(26)-\log(38)]\}\times10^{-18}$$ $$=\{\text{antilog}[0.8215+1.5563-1.4150-1.5798]\}\times10^{-18}$$ $$=\{\text{antilog}[-0.6170]\}\times10^{-18}$$ $$=\{\text{antilog}[\overline{1}.3830]\}\times10^{-18}$$ $$=2.416\times10^{-1}\times10^{-18}=2.416\times10^{-19}\text{ J}$$ $$\therefore(K.E.)_{max}(\text{in eV})=\dfrac{2.416\times10^{-19}}{1.6\times10^{-19}}=1.509\text{ eV}$$ Ans: Maximum kinetic energy of the photoelectron emitted is 1.509 eV.

15. 15.↑ back to the question

    Given: $V_0=1.8$ V, $e=1.6\times10^{-19}$ C, $\lambda=2200$ Å To find: Maximum kinetic energy (K.E.)$_{max}$ Formula: $(K.E.)_{max}=eV_0$ Calculation: Using formula, $(K.E.)_{max}=1.6\times10^{-19}\times1.8$ $\therefore(K.E)_{max}=2.88\times10^{-19}$ J Ans: Maximum kinetic energy of emitted photoelectron is $2.88\times10^{-19}$ J.

16. 16.↑ back to the question

    Given: $\phi_0=2.3$ eV, $\lambda=6800$ Å $=6.8\times10^{-7}$ m, $c=3\times10^8$ m/s, $h=6.63\times10^{-34}$ Js To find: i. Threshold frequency ($\nu_0$) ii. Incident frequency ($\nu$) iii. If emission of photoelectrons possible Formulae: i. $\phi_0=h\nu_0$ ii. $\nu=\dfrac{c}{\lambda}$ Calculation: From formula (i), $$\nu_0=\dfrac{\phi_0}{h}=\dfrac{2.3\times1.6\times10^{-19}}{6.63\times10^{-34}}=\dfrac{3.68\times10^{15}}{6.63}=\dfrac{36.8}{6.63}\times10^{14}$$ $$=\{\text{antilog}[\log(36.8)-\log(6.63)]\}\times10^{14}$$ $$=\{\text{antilog}(1.5658-0.8215)\}\times10^{14}$$ $$=\{\text{antilog}(0.7443)\}\times10^{14}$$ $$=5.550\times10^{14}\text{ Hz}$$ From formula (ii) $$\nu=\dfrac{3\times10^8}{6.8\times10^{-7}}=\dfrac{30\times10^7}{6.8\times10^{-7}}$$ $$=\{\text{antilog}[\log(30)-\log(6.8)]\}\times10^{14}$$ $$=\{\text{antilog}(1.4771-0.8325)\}\times10^{14}$$ $$=\{\text{antilog}(0.6446)\}\times10^{14}$$ $$=4.412\times10^{14}\text{ Hz}$$ Since incident frequency is less than threshold frequency i.e., $\nu<\nu_0$, photoelectric emission will not be observed. Ans: i. Threshold frequency $\nu_0$ is $5.55\times10^{14}$ Hz. ii. Frequency of incident light is $4.412\times10^{14}$ Hz. iii. Emission of photoelectrons is not possible.

17. 17.↑ back to the question

    Given: $\lambda=3000$ Å $=3\times10^{-7}$ m, $m_e=9.1\times10^{-31}$ kg, $\phi_0=2.3$ eV, $h=6.63\times10^{-34}$ Js, $c=3\times10^8$ m/s To find: Maximum velocity ($v_{max}$) Formulae: i. $(K.E)_{max}=\dfrac{hc}{\lambda}-\phi_0$ ii. $v_{max}=\sqrt{\dfrac{2(K.E.)_{max}}{m_e}}$ Calculation: From formula (i), $$(K.E.)_{max}=\left[\dfrac{6.63\times10^{-34}\times3\times10^8}{3\times10^{-7}}-(2.3\times1.6\times10^{-19})\right]$$ $$=(6.63\times10^{-19})-(3.68\times10^{-19})=2.95\times10^{-19}$$ Also, $\dfrac{1}{2}m_ev_{max}^2=(K.E.)_{max}$ From formula (ii), $$v_{max}=\sqrt{\dfrac{2\times2.95\times10^{-19}}{9.1\times10^{-31}}}=\sqrt{\dfrac{5.90\times10^{12}}{9.1}}=\sqrt{\dfrac{590}{9.1}}\times10^5$$ $$=\left\{\text{antilog}\dfrac{1}{2}[\log(590)-\log(9.1)]\right\}\times10^5$$ $$=\left\{\text{antilog}\left[\dfrac{1}{2}(2.7709-0.9590)\right]\right\}\times10^5$$ $$=\left\{\text{antilog}\left[\dfrac{1}{2}(1.8119)\right]\right\}\times10^5$$ $$=\{\text{antilog}(0.9059)\}\times10^5=8.052\times10^5\text{ m/s}$$ Ans: The maximum velocity of electron is $8.052\times10^5$ m/s

18. 18.↑ back to the question

    Given: $(\phi_0)_P=2.25$ eV $=2.25\times1.6\times10^{-19}$ J $=3.6\times10^{-19}$ J, $(\phi_0)_C=2.14$ eV $=2.14\times1.6\times10^{-19}$ J $=3.424\times10^{-19}$ J, $\lambda=5180$ Å $=5.18\times10^{-7}$ m To find: Will the photoelectric effect occur for potassium and caesium with $\lambda=5180$ Å Formula: $\phi_0=\dfrac{hc}{\lambda_0}$ Calculation: From formula, $$(\phi_0)_P=\dfrac{hc}{(\lambda_0)_P}$$ $$\therefore(\lambda_0)_P=\dfrac{hc}{(\phi_0)_P}=\dfrac{6.63\times10^{-34}\times3\times10^8}{3.6\times10^{-19}}=\dfrac{6.63}{1.2}\times10^{-7}=5.525\times10^{-7}\text{ m}=5525\text{ \text{Å}}$$ Similarly, $$(\lambda_0)_C=\dfrac{hc}{(\phi_0)_C}=\dfrac{6.63\times10^{-34}\times3\times10^8}{3.424\times10^{-19}}=5.809\times10^{-7}\text{ m}=5809\text{ \text{Å}}$$ As incident wavelength ($\lambda$) is smaller than threshold wavelength for potassium $(\lambda_0)_P$ and for caesium $(\lambda_0)_C$ i.e., $\lambda<(\lambda_0)_P<(\lambda_0)_C$. Photoelectric effect is possible for both of them. Ans: For $\lambda=5180$ Å wavelength, both potassium and caesium will exhibit photoelectric emission.

19. 19.↑ back to the question

    Given: $\phi_0=3$ eV $=3\times1.6\times10^{-19}$ J, $h=6.63\times10^{-34}$ J-s, $e=1.6\times10^{-19}$ C, $c=3\times10^8$ m/s. To find: Threshold wavelength ($\lambda_0$) Formula: $\lambda_0=\dfrac{hc}{\phi_0}$ Calculation: From formula, $$\lambda_0=\dfrac{6.63\times10^{-34}\times3\times10^8}{3\times1.6\times10^{-19}}=4.14\times10^{-7}\text{ m}$$ Ans: Threshold wavelength of the metal is $4.14\times10^{-7}$ m.

20. 20.↑ back to the question

    Given: $E=6.63$ J, $\nu=10^{14}$ Hz, $h=6.63\times10^{-34}$ Js. To find: Number of photons (n) Formula: $n=\dfrac{E}{h\nu}$ Calculation: Using formula, $$n=\dfrac{6.63}{6.63\times10^{-34}\times10^{14}}=10^{20}$$ Ans: The number of photons emitted in the radiation are $10^{20}$.

21. 21.↑ back to the question

    Given: $\phi_0=2.2$ eV $=2.2\times1.6\times10^{-19}$ J, $\lambda=5000$ Å $=5000\times10^{-10}$ m, $h=6.63\times10^{-34}$ J.s, $c=3\times10^8$ m/s To find: i. Threshold frequency ($\nu_0$) ii. Incident frequency ($\nu$) iii. If photoemission is possible? Formulae: i. $\phi_0=h\nu_0$ ii. $c=\nu\lambda$ Calculation: Using formula (i), $2.2\times1.6\times10^{-19}=6.63\times10^{-34}\times\nu_0$ $$\nu_0=\dfrac{2.2\times1.6\times10^{-19}}{6.63\times10^{-34}}$$ $$=\{\text{antilog}[\log(2.2)+\log(1.6)-\log(6.63)]\}\times10^{15}$$ $$=\{\text{antilog}[0.3424+0.2041-0.8215]\}\times10^{15}$$ $$=\{\text{antilog}(\overline{1}.6250)\}\times10^{15}$$ $$=0.5309\times10^{15}$$ $\nu_0=5.31\times10^{14}$ Hz Using formula (ii), $3\times10^8=\nu\times5000\times10^{-10}$ $$\nu=\dfrac{3\times10^8}{5000\times10^{-10}}$$ $\nu=6\times10^{14}$ Hz As, $\nu>\nu_0$, there will be emission of photoelectrons. Ans: i. Threshold frequency is $5.31\times10^{14}$ Hz ii. Incident frequency is $6\times10^{14}$ Hz iii. There will be emission of photoelectrons.

22. 22.↑ back to the question

    Given: $\phi_0=3.84\times10^{-19}$ J, $\lambda=5000$ Å $=5000\times10^{-10}$ m Energy of incident radiation $(E)=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^8}{5000\times10^{-10}}=3.978\times10^{-19}$ J Here, $E>\phi_0$ Since E is greater than $\phi_0$, photoelectric emission will occur. Ans: Yes, photoelectric emission will occur.

23. 23.↑ back to the question

    Given: $\phi_0=4.2$ eV $=4.2\times1.6\times10^{-19}$ J To find: Threshold wavelength ($\lambda_0$) Formula: $\phi_0=\dfrac{hc}{\lambda_0}$ Calculation: From formula, $$\lambda_0=\dfrac{hc}{\phi_0}=\dfrac{6.63\times10^{-34}\times3\times10^8}{4.2\times1.6\times10^{-19}}=2.9598\times10^{-7}\text{ m}=2959.8\text{ \text{Å}}$$ Ans: The threshold wavelength is 2959.8 Å.

24. 24.↑ back to the question

    Maximum kinetic energy of a photoelectron is given as, $$KE_{max}=eV_0=1.6\times10^{-19}\times1.5=2.4\times10^{-19}\text{ J}$$