MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterMonochromatic light of wavelength $500\text{ nm}$ is incident on a photoelectric surface. The work function of the metal is $2.14\text{ eV}$. (a) Calculate the energy of the incident photon in electron-volts ($\text{eV}$). (b) Determine the maximum kinetic energy of the emitted photoelectrons. (c) Calculate the stopping potential for this surface. (d) Find the threshold frequency for the metal.

Given Data:

• Wavelength of incident light, $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$
• Work function of the metal, $\phi_0 = 2.14\text{ eV}$
• Planck's constant, $h = 6.63 \times 10^{-34}\text{ J s}$
• Speed of light in vacuum, $c = 3 \times 10^8\text{ m/s}$
• Charge of electron, $e = 1.6 \times 10^{-19}\text{ C}$

(a) Calculation of Energy of the Incident Photon ($E$):\nThe energy of a photon is given by the formula:

$$E = \frac{hc}{\lambda}$|\nSubstituting the given values: $$E = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3 \times 10^8\text{ m/s})}{500 \times 10^{-9}\text{ m}}$$ $$E = \frac{1.989 \times 10^{-25}}{500 \times 10^{-9}} = 3.978 \times 10^{-19}\text{ J}$$\nConverting joules into electron-volts ($\text{eV}$) by dividing by $1.6 \times 10^{-19}\text{ J/eV}$: $$E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.486\text{ eV} \approx 2.49\text{ eV}$$

(b) Determination of Maximum Kinetic Energy ($K_{max}$):\nAccording to Einstein's photoelectric equation:

$$K_{max} = E - \phi_0$$\nSubstituting the values obtained: $$K_{max} = 2.49\text{ eV} - 2.14\text{ eV} = 0.35\text{ eV}$$\nConverting into joules: $$K_{max} = 0.35 \times 1.6 \times 10^{-19}\text{ J} = 5.6 \times 10^{-20}\text{ J}$$

(c) Calculation of Stopping Potential ($V_0$):\nThe stopping potential is related to maximum kinetic energy by:

$$eV_0 = K_{max}$$ $$V_0 = \frac{K_{max}}{e}$$\nSubstituting the values in $\text{eV}$ and $e$: $$V_0 = \frac{0.35\text{ eV}}{1\text{ e}} = 0.35\text{ V}$$

(d) Calculation of Threshold Frequency ($\nu_0$):\nThe work function is related to threshold frequency by:

$$\phi_0 = h\nu_0$$ $$\nu_0 = \frac{\phi_0}{h}$$\nFirst, convert work function $\phi_0$ into joules: $$\phi_0 = 2.14 \times 1.6 \times 10^{-19}\text{ J} = 3.424 \times 10^{-19}\text{ J}$$\nNow, calculate $\nu_0$: $$\nu_0 = \frac{3.424 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}}$$ $$\nu_0 \approx 5.16 \times 10^{14}\text{ Hz}$$

Final Answers:

(a) Energy of incident photon = $2.49\text{ eV}$ (b) Maximum kinetic energy = $0.35\text{ eV} = 5.6 \times 10^{-20}\text{ J}$ (c) Stopping potential = $0.35\text{ V}$ (d) Threshold frequency = $5.16 \times 10^{14}\text{ Hz}$