When light having a wavelength of 497 nm strikes a metal coating of an emitter, an elector with kinetic energy of 2.52 x 10^-19 is ejected. What is the threshold frequency for this metal?
508 cm fly
To find the threshold frequency for the metal, we need to use the equation that relates the energy of a photon to its frequency:
E = h * f
where E is the energy of the photon, h is the Planck constant (6.62607015 × 10^-34 J·s), and f is the frequency of the photon.
We can rearrange this equation to solve for the frequency:
f = E / h
The threshold frequency corresponds to the minimum frequency required for an electron to be ejected. This occurs when all the energy of the incident photon is used to overcome the binding energy (work function) and free the electron. In this case, the kinetic energy of the ejected electron is equal to the energy of the incident photon.
Given that the kinetic energy of the ejected electron is 2.52 x 10^-19 J, we can use this value to find the frequency:
f_threshold = (2.52 x 10^-19 J) / (6.62607015 × 10^-34 J·s)
Calculating this:
f_threshold ≈ 3.80 x 10^14 Hz
Now, to find the threshold frequency in terms of wavelength, we can use the equation that relates frequency and wavelength:
c = λ * f
where c is the speed of light (approximately 3.00 x 10^8 m/s), λ is the wavelength, and f is the frequency.
Rearranging this equation to solve for wavelength:
λ = c / f_threshold
Substituting the values:
λ ≈ (3.00 x 10^8 m/s) / (3.80 x 10^14 Hz)
Calculating this:
λ ≈ 7.89 x 10^-7 m
Therefore, the threshold frequency for this metal is approximately 3.80 x 10^14 Hz or, equivalently, the threshold wavelength is approximately 7.89 x 10^-7 m.