MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterMonochromatic radiation of wavelength $6400 \text{ \u00c5}$ is incident on a photosensitive surface. If the work function of the metal is $2.0 \text{ eV}$, calculate: (i) The energy of an incident photon in eV. (ii) The kinetic energy of the emitted photoelectrons. (iii) The stopping potential. (Given: $h = 6.63 \times 10^{-34} \text{ J s}$, $c = 3 \times 10^8 \text{ m s}^{-1}$, $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$)

Given Data:

• Wavelength of incident radiation ($\lambda$) = $6400 \text{ \u00c5} = 6400 \times 10^{-10} \text{ m} = 6.4 \times 10^{-7} \text{ m}$
• Work function of metal ($\phi_0$) = $2.0 \text{ eV}$
• Planck's constant ($h$) = $6.63 \times 10^{-34} \text{ J s}$
• Speed of light ($c$) = $3 \times 10^8 \text{ m s}^{-1}$
• Charge of electron ($e$) = $1.6 \times 10^{-19} \text{ C}$-

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

$$E = \frac{hc}{\lambda}$| \nSubstituting the values: $$E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m s}^{-1})}{6.4 \times 10^{-7} \text{ m}}$$ $$E = \frac{19.89 \times 10^{-26}}{6.4 \times 10^{-7}}$$ $$E = 3.1078 \times 10^{-19} \text{ J}$| \nConverting energy into electron-volts (eV): $$E = \frac{3.1078 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} \approx 1.94 \text{ eV}$|

(ii) Calculation of Kinetic Energy ($K_{max}$) of Emitted Photoelectrons\nUsing Einstein's photoelectric equation:

$$K_{max} = E - \phi_0$$ \nSince the incident photon energy ($1.94 \text{ eV}$) is less than the work function ($2.0 \text{ eV}$), photoelectric emission will not take place.\nTherefore, $K_{max} = 0 \text{ eV}$. (Note: If the problem assumes emission is possible for calculation purposes, let us check the threshold condition. Since $E < \phi_0$, no photoelectrons are emitted. However, if the question intends standard evaluation, we state that kinetic energy is zero as emission does not occur.)

(iii) Calculation of Stopping Potential ($V_0$)\nSince $K_{max} = 0$, the stopping potential is also:

$$V_0 = 0 \text{ V}$|

Final Answer:

(i) Energy of the incident photon = $1.94 \text{ eV}$ (ii) Kinetic energy of photoelectrons = $0 \text{ eV}$ (No emission occurs as $E < \phi_0$) (iii) Stopping potential = $0 \text{ V}$