Understanding Photoelectric Emission and Threshold Wavelength

Photoelectric emission is a phenomenon where electrons are ejected from a material surface when light of sufficient frequency shines on it. The minimum energy required to remove an electron from the surface of a material is called the work function ($\phi_0$). Each material has a specific work function. Light energy is carried by photons, with each photon having energy given by $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the light. When a photon strikes the surface, it can transfer its energy to an electron. If the photon's energy is greater than or equal to the work function ($\phi_0$), the electron can be ejected. The threshold wavelength ($\lambda_0$) is the maximum wavelength of light that can cause photoelectric emission from a surface. This corresponds to the minimum frequency ($\nu_0$) required, such that the photon energy $h\nu_0$ is equal to the work function $\phi_0$. Since the speed of light $c = \nu\lambda$, the threshold frequency $\nu_0 = c/\lambda_0$. Therefore, the relationship between work function ($\phi_0$) and threshold wavelength ($\lambda_0$) is: $$ \phi_0 = h\nu_0 = \frac{hc}{\lambda_0} $$ We are given the work function for an Aluminium surface as $\phi_0 = 4.2 \text{ eV}$. We need to find the threshold wavelength ($\lambda_0$).

Calculating the Threshold Wavelength

To calculate the threshold wavelength using the formula $\lambda_0 = \frac{hc}{\phi_0}$, we need the values for Planck's constant ($h$) and the speed of light ($c$). We also need the work function in units consistent with $h$ and $c$, typically Joules (J), if $h$ is in J s and $c$ is in m/s. Standard values for the constants are:

• Planck's constant, $h = 6.626 \times 10^{-34} \text{ J s}$
• Speed of light in vacuum, $c = 3.00 \times 10^8 \text{ m/s}$

The work function is given in electron volts (eV). We need to convert it to Joules. The conversion factor is $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. Convert the work function from eV to Joules: $$ \phi_0 = 4.2 \text{ eV} \times \left(1.602 \times 10^{-19} \frac{\text{J}}{\text{eV}}\right) $$ $$ \phi_0 = 6.7284 \times 10^{-19} \text{ J} $$ Now, we can calculate the threshold wavelength using the formula: $$ \lambda_0 = \frac{hc}{\phi_0} $$ Substitute the values: $$ \lambda_0 = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m/s})}{6.7284 \times 10^{-19} \text{ J}} $$ $$ \lambda_0 = \frac{19.878 \times 10^{-26} \text{ J m}}{6.7284 \times 10^{-19} \text{ J}} $$ $$ \lambda_0 \approx 2.9542 \times 10^{-7} \text{ m} $$ The options are given in Angstroms (Å). We need to convert the result from meters to Angstroms. The conversion factor is $1 \text{ Å} = 10^{-10} \text{ m}$, or $1 \text{ m} = 10^{10} \text{ Å}$. Convert the threshold wavelength from meters to Angstroms: $$ \lambda_0 \approx (2.9542 \times 10^{-7} \text{ m}) \times \left(\frac{10^{10} \text{ Å}}{1 \text{ m}}\right) $$ $$ \lambda_0 \approx 2.9542 \times 10^{(-7+10)} \text{ Å} $$ $$ \lambda_0 \approx 2.9542 \times 10^{3} \text{ Å} $$ $$ \lambda_0 \approx 2954.2 \text{ Å} $$ Comparing this value with the given options:

• 2959 Å
• 4200 Å
• 4736 Å
• 5890 Å

Our calculated value of approximately 2954.2 Å is closest to 2959 Å. The slight difference might be due to using slightly different values for constants like $h$, $c$, or the eV to Joule conversion factor in the options' calculation. Often, in atomic physics calculations, the product $hc$ is used directly. Using $hc \approx 1242 \text{ eV nm}$ or $hc \approx 12420 \text{ eV Å}$ gives a result very close to 2959 Å: $$ \lambda_0 = \frac{12420 \text{ eV Å}}{4.2 \text{ eV}} \approx 2957.14 \text{ Å} $$ This confirms that the option 2959 Å is the intended answer based on typical constant values used in such problems.

Result

The threshold wavelength for photoelectric emission from the Aluminium surface with a work function of 4.2 eV is approximately 2954.2 Å, which is closest to 2959 Å among the given options.

Parameter	Value	Unit
Work Function ($\phi_0$)	4.2	eV
Work Function ($\phi_0$)	$6.7284 \times 10^{-19}$	J
Planck's Constant ($h$)	$6.626 \times 10^{-34}$	J s
Speed of Light ($c$)	$3.00 \times 10^8$	m/s
Calculated Threshold Wavelength ($\lambda_0$)	$2.9542 \times 10^{-7}$	m
Calculated Threshold Wavelength ($\lambda_0$)	2954.2	Å

Revision Table: Key Concepts

Term	Definition	Formula (related to work function)
Work Function ($\phi_0$)	Minimum energy to remove an electron from a surface.	Given for the material.
Threshold Frequency ($\nu_0$)	Minimum light frequency for photoelectric emission.	$\phi_0 = h\nu_0$
Threshold Wavelength ($\lambda_0$)	Maximum light wavelength for photoelectric emission.	$\phi_0 = \frac{hc}{\lambda_0}$

Additional Information: Photoelectric Effect

The photoelectric effect is a key concept in quantum physics, first explained by Albert Einstein, building on Max Planck's quantum hypothesis. Key aspects include:

• Photon Nature of Light: Light behaves as packets of energy called photons.
• Energy Quanta: Each photon of frequency $\nu$ carries a fixed energy $E = h\nu$.
• One-to-One Interaction: An electron absorbs energy from a single photon.
• Kinetic Energy of Emitted Electrons: If a photon's energy ($h\nu$) is greater than the work function ($\phi_0$), the excess energy becomes the kinetic energy ($K_{max}$) of the emitted electron: $K_{max} = h\nu - \phi_0$. This is Einstein's photoelectric equation.
• Threshold Concept: If the photon energy $h\nu$ is less than the work function $\phi_0$, no electron is emitted, regardless of the intensity of the light. This explains the existence of a threshold frequency and threshold wavelength.
• Instantaneous Emission: Photoelectric emission is nearly instantaneous, even at very low light intensities, as energy absorption is from individual photons rather than accumulated over time.

This effect provided strong evidence for the particle nature of light and was a crucial step in the development of quantum mechanics.