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Question

The photoelectric threshold wavelength of silver is $3250 	imes {10^{ - 10}}{m{ m}}{m{.}}$The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 	imes {10^{ - 10}}{m{ms}}$(Given $h = 4.14 	imes {10^{ - 15}}{m{ eVs \;and \;}}c = 3 	imes {10^8}{m{ m}}{{m{s}}^{ - 1}}$)

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Photoelectric Effect
	**Related Topics**: Work Function, Kinetic Energy of Photoelectrons, Energy of Photon
	**Prerequisites**: Understanding of electromagnetic radiation, Planck's constant, and kinetic energy.
## SOLUTION PATHWAY
**Given Information**:
	* Threshold wavelength ($λ_0$) = $3250 	imes 10^{-10}$ m
	* Incident wavelength ($λ$) = $2536 	imes 10^{-10}$ m
	* Planck's constant (h) = $4.14 	imes 10^{-15}$ eVs
	* Speed of light (c) = $3 	imes 10^8$ m/s

**Method & Steps**:
	1. Calculate the work function (Φ) using the threshold wavelength:
		Reasoning: The work function is the minimum energy required to eject an electron, which corresponds to the threshold frequency/wavelength.
		Formula/Rule:  $Φ = \frac{hc}{λ_0}$
		Working: $Φ = \frac{(4.14 	imes 10^{-15} 	ext{ eV s}) (3 	imes 10^8 	ext{ m/s})}{3250 	imes 10^{-10} 	ext{ m}} = 3.828 	ext{ eV}$
	2. Calculate the energy of the incident photon (E):
		Reasoning: The energy of the incident photon is given by the formula E = hc/λ.
		Formula/Rule: $E = \frac{hc}{λ}$
		Working: $E = \frac{(4.14 	imes 10^{-15} 	ext{ eV s}) (3 	imes 10^8 	ext{ m/s})}{2536 	imes 10^{-10} 	ext{ m}} = 4.896 	ext{ eV}$
	3. Calculate the kinetic energy (KE) of the emitted electron:
		Reasoning: The kinetic energy of the emitted electron is the difference between the energy of the incident photon and the work function.
		Formula/Rule: $KE = E - Φ$
		Working: $KE = 4.896 	ext{ eV} - 3.828 	ext{ eV} = 1.068 	ext{ eV}$
	4. Convert the kinetic energy from eV to Joules: 
		Reasoning: To find the velocity, we need the kinetic energy in Joules.
		Formula/Rule: 1 eV = $1.602 	imes 10^{-19}$ J
		Working: $KE = 1.068 	ext{ eV} 	imes 1.602 	imes 10^{-19} 	ext{ J/eV} = 1.711 	imes 10^{-19} 	ext{ J}$
	5. Calculate the velocity (v) of the emitted electron:
		Reasoning: Using the kinetic energy formula $KE = \frac{1}{2}mv^2$, we can solve for v.
		Formula/Rule: $v = \sqrt{\frac{2KE}{m}}$ where the mass of an electron (m) is $9.11 	imes 10^{-31}$ kg.
		Working: $v = \sqrt{\frac{2 	imes 1.711 	imes 10^{-19} 	ext{ J}}{9.11 	imes 10^{-31} 	ext{ kg}}}  \approx 6.13 	imes 10^5 	ext{ m/s} \approx 0.6 	imes 10^6 	ext{ m/s}$
## OPTION ANALYSIS
[A] :x: The calculated velocity is approximately 0.6 x 10^6 m/s, not 0.3 x 10^6 m/s
[B] :x: The calculated velocity is approximately 0.6 x 10^6 m/s, not 6 x 10^5 m/s
[C] ✓: The calculated velocity is approximately 0.6 x 10^6 m/s
[D] :x: The calculated velocity is approximately 0.6 x 10^6 m/s, not 61 x 10^3 m/s.
## LEARNING AIDS
	**Common Mistakes**:
		1. **Unit Conversion**: Forgetting to convert eV to Joules when calculating kinetic energy can lead to incorrect results.
		2. **Incorrect Formula**: Using the wrong relationship between energy and wavelength or between kinetic energy and velocity.
	**Quick Tips**:
		1. **Photoelectric Equation**: Remember the equation $KE = E - Φ$ which is the core principle behind the photoelectric effect.
		2. **Unit Consistency**: Always make sure all units are consistent before performing calculations (e.g., Joules for energy, meters per second for velocity).

NAGA SAI · Answer

$ \approx 0.3 	imes {10^6}{m{ m}}{{m{s}}^{ - 1}}$

NAGA SAI · Answer

$ \approx 6 	imes {10^5}{m{ m}}{{m{s}}^{ - 1}}$

NAGA SAI · Answer

The correct Option is "$ \approx 0.6 	imes {10^6}{m{ m}}{{m{s}}^{ - 1}}$"

NAGA SAI · Answer

$ \approx 61 	imes {10^3}{m{ m}}{{m{s}}^{ - 1}}$

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