Monochromatic light of wavelength $500 ext{ nm}$ is incident on a photoelectric surface. Work function of the metal is $2.13 ext{ eV}$. Calculate: (i) The maximum kinetic energy of emitted photoelectrons in Joules and eV, (ii) The stopping potential, and (iii) The threshold frequency for the metal surface. (Take $h = 6.63 imes 10^{-34} ext{ J s}$, $c = 3 imes 10^8 ext{ m/s}$).

Q: Monochromatic light of wavelength $500\text{ nm}$ is incident on a photoelectric surface. Work function of the metal is $2.13\text{ eV}$. Calculate: (i) The maximum kinetic energy of emitted photoelectrons in Joules and eV, (ii) The stopping potential, and (iii) The threshold frequency for the metal surface. (Take $h = 6.63 \times 10^{-34}\text{ J s}$, $c = 3 \times 10^8\text{ m/s}$).

Given Data: Wavelength of incident light, $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$ Work function of the metal, $\phi0 = 2.13\text{ eV} = 2.13 \times 1.6 \times 10^{-19}\text{ J} = 3.408 \times 10^{-19}\text{ J}$ Planck's constant, $h = 6.63 \times 10^{-34}\text{ J s}$ Speed of light, $c = 3 \times 10^8\text{ m/s}$ Charge of electron, $e = 1.6 \times 10^{-19}\text{ C}$ --- Step (i): Calculate the maximum kinetic energy ($K{max}$) of emitted photoelectrons\nFirst, calculate the energy of the incident photon ($E$): $$E = \frac{hc}{\lambda}$$\nSubstitute the values: $$E = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3 \times 10^8\text{ m/s})}{500 \times 10^{-9}\text{ m}}$$ $$E = \frac{1.989 \times 10^{-25}}{500 \times 10^{-9}} = 3.978 \times 10^{-19}\text{ J}$| \nConvert photon energy into electron-volts (eV): $$E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.486\text{ eV}$$ \nNow, use Einstein's photoelectric equation to find $K{max}$: $$K{max} = E - \phi0$$ $$K{max} = 2.486\text{ eV} - 2.13\text{ eV} = 0.356\text{ eV}$$ \nIn Joules: $$K{max} = 0.356 \times 1.6 \times 10^{-19}\text{ J} = 5.696 \times 10^{-20}\text{ J}$$ --- Step (ii): Calculate the stopping potential ($V0$)\nThe relation between maximum kinetic energy and stopping potential is: $$K{max} = eV0$$ $$V0 = \frac{K{max}}{e}$$ $$V0 = \frac{0.356 \times 1.6 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 0.356\text{ V}$$ --- Step (iii): Calculate the threshold frequency ($\nu0$)\nThe relation between work function and threshold frequency is: $$\phi0 = h\nu0$$ $$\nu0 = \frac{\phi0}{h}$$\nSubstitute the values: $$\nu0 = \frac{3.408 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}}$$ $$\nu0 = 0.514 \times 10^{15}\text{ Hz} = 5.14 \times 10^{14}\text{ Hz}$$ --- Final Answer: (i) Maximum Kinetic Energy: $5.696 \times 10^{-20}\text{ J}$ (or $0.356\text{ eV}$) (ii) Stopping Potential: $0.356\text{ V}$ (iii) Threshold Frequency: $5.14 \times 10^{14}\text{ Hz}$

Question

Monochromatic light of wavelength $500	ext{ nm}$ is incident on a photoelectric surface. Work function of the metal is $2.13	ext{ eV}$. Calculate: (i) The maximum kinetic energy of emitted photoelectrons in Joules and eV, (ii) The stopping potential, and (iii) The threshold frequency for the metal surface. (Take $h = 6.63 	imes 10^{-34}	ext{ J s}$, $c = 3 	imes 10^8	ext{ m/s}$).

Accepted Answer

Given Data: Wavelength of incident light, $\lambda = 500	ext{ nm} = 500 	imes 10^{-9}	ext{ m}$ Work function of the metal, $\phi0 = 2.13	ext{ eV} = 2.13 	imes 1.6 	imes 10^{-19}	ext{ J} = 3.408 	imes 10^{-19}	ext{ J}$ Planck's constant, $h = 6.63 	imes 10^{-34}	ext{ J s}$ Speed of light, $c = 3 	imes 10^8	ext{ m/s}$ Charge of electron, $e = 1.6 	imes 10^{-19}	ext{ C}$ --- Step (i): Calculate the maximum kinetic energy ($K{max}$) of emitted photoelectrons
First, calculate the energy of the incident photon ($E$): $$E = \frac{hc}{\lambda}$$
Substitute the values: $$E = \frac{(6.63 	imes 10^{-34}	ext{ J s}) 	imes (3 	imes 10^8	ext{ m/s})}{500 	imes 10^{-9}	ext{ m}}$$ $$E = \frac{1.989 	imes 10^{-25}}{500 	imes 10^{-9}} = 3.978 	imes 10^{-19}	ext{ J}$| 
Convert photon energy into electron-volts (eV): $$E = \frac{3.978 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 2.486	ext{ eV}$$ 
Now, use Einstein's photoelectric equation to find $K{max}$: $$K{max} = E - \phi0$$ $$K{max} = 2.486	ext{ eV} - 2.13	ext{ eV} = 0.356	ext{ eV}$$ 
In Joules: $$K{max} = 0.356 	imes 1.6 	imes 10^{-19}	ext{ J} = 5.696 	imes 10^{-20}	ext{ J}$$ --- Step (ii): Calculate the stopping potential ($V0$)
The relation between maximum kinetic energy and stopping potential is: $$K{max} = eV0$$ $$V0 = \frac{K{max}}{e}$$ $$V0 = \frac{0.356 	imes 1.6 	imes 10^{-19}	ext{ J}}{1.6 	imes 10^{-19}	ext{ C}} = 0.356	ext{ V}$$ --- Step (iii): Calculate the threshold frequency ($
u0$)
The relation between work function and threshold frequency is: $$\phi0 = h
u0$$ $$
u0 = \frac{\phi0}{h}$$
Substitute the values: $$
u0 = \frac{3.408 	imes 10^{-19}	ext{ J}}{6.63 	imes 10^{-34}	ext{ J s}}$$ $$
u0 = 0.514 	imes 10^{15}	ext{ Hz} = 5.14 	imes 10^{14}	ext{ Hz}$$ --- Final Answer: (i) Maximum Kinetic Energy: $5.696 	imes 10^{-20}	ext{ J}$ (or $0.356	ext{ eV}$) (ii) Stopping Potential: $0.356	ext{ V}$ (iii) Threshold Frequency: $5.14 	imes 10^{14}	ext{ Hz}$

Given Data:

Step (i): Calculate the maximum kinetic energy ($K_{max}$) of emitted photoelectrons\nFirst, calculate the energy of the incident photon ($E$):

Step (ii): Calculate the stopping potential ($V_0$)\nThe relation between maximum kinetic energy and stopping potential is:

Step (iii): Calculate the threshold frequency ($\nu_0$)\nThe relation between work function and threshold frequency is:

Final Answer: