MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterMonochromatic light of wavelength $500\text{ nm}$ is incident on a certain metal surface. If the work function of the metal is $2.1\text{ eV}$, calculate: (a) the energy of the incident photon in electron-volts, (b) the maximum kinetic energy of the emitted photoelectrons, and (c) the stopping potential. Given $h = 6.63 \times 10^{-34}\text{ J s}$, $c = 3 \times 10^8\text{ m/s}$, and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$.

Step-by-Step Solution:

Given data:

• Wavelength of incident light, $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m} = 5 \times 10^{-7}\text{ m}$
• Work function of metal, $\phi_0 = 2.1\text{ eV}$
• Planck's constant, $h = 6.63 \times 10^{-34}\text{ J s}$
• Speed of light, $c = 3 \times 10^8\text{ m/s}$
• Conversion factor: $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$

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

$$E = \frac{hc}{\lambda}$$\nSubstituting the given values: $$E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3 \times 10^8\text{ m/s}}{5 \times 10^{-7}\text{ m}}$$ $$E = \frac{1.989 \times 10^{-25}}{5 \times 10^{-7}} = 3.978 \times 10^{-19}\text{ J}$$ \nConverting energy into electron-volts (eV): $$E = \frac{3.978 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} = 2.486\text{ eV}$$

(b) Maximum kinetic energy of the emitted photoelectrons ($K_{max}$)\nUsing Einstein's photoelectric equation:

$$K_{max} = E - \phi_0$$\nSubstituting the values in eV: $$K_{max} = 2.486\text{ eV} - 2.1\text{ eV} = 0.386\text{ eV}$$ \nConverting maximum kinetic energy into Joules: $$K_{max} = 0.386 \times 1.6 \times 10^{-19}\text{ J} = 6.176 \times 10^{-20}\text{ J}$$

(c) Stopping potential ($V_0$)\nThe relationship between maximum kinetic energy and stopping potential is:

$$K_{max} = e V_0$$\nTherefore, the stopping potential $V_0$ is: $$V_0 = \frac{K_{max}}{e}$|\nSubstituting the value of $K_{max}$ in eV and electron charge $e$: $$V_0 = \frac{0.386\text{ eV}}{e} = 0.386\text{ V}$$

Final Answer:

(a) Energy of the incident photon = $2.49\text{ eV}$ (b) Maximum kinetic energy = $0.386\text{ eV}$ (or $6.18 \times 10^{-20}\text{ J}$) (c) Stopping potential = $0.386\text{ V}$