Question

Silver has a work function, ϕ, of 4.64 ev. If a sample of silver is bombarded with 215 nm photons, what will be the wavelength (in meters) of the photo-ejected electrons?

Answer 1

The wavelength of photo-ejected electrons from silver bombarded with 215 nm photons can be calculated using the conservation of energy in the photoelectric effect, the energy of the incident photons, and the De Broglie wavelength formula.

Step-by-step explanation:

The question asks about the photoelectric effect as it pertains to a sample of silver being bombarded with 215 nm photons and the wavelength of the ejected electrons. To solve for the wavelength of the photo-ejected electrons, we use the equation derived from the conservation of energy in the photoelectric effect:

• 
  
• E_photon = φ + E_kinetic

Where φ is the work function of silver, which is given as 4.64 eV, and E_photon is the energy of the incoming photons, which can be calculated using the equation:

• 
  
• E_photon = (hc)/λ

Here, h is Planck's constant (4.135667696 x 10^-15 eV•s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the incident photons (215 nm).

First, we convert the wavelength from nanometers to meters to maintain SI units throughout the calculation:

• 
  
• 215 nm = 215 x 10^-9 m

Next, calculate the energy of the incident photons (E_photon).

Now, to find the maximum kinetic energy (E_kinetic) of the ejected electrons:

• 
  
• E_kinetic = E_photon - φ

Using the kinetic energy, we can find the velocity of the ejected electrons using the formula:

• 
  
• E_kinetic = 0.5 x m_e x v^2

Where m_e is the mass of an electron (9.10938356 x 10^-31 kg).

After calculating the velocity (v), we use the De Broglie wavelength formula to find the wavelength (λ_electron) of the ejected electrons:

• 
  
• λ_electron = h/m_e v

The calculated wavelength (in meters) will be the final answer to the student's question.