Photons of energy 12eV are incident on a metal. It is found that current flows from the metal until a stopping potential of 8.0V is applied.
What would happen if the wavelength of the incident photons was tripled?
Would the energy be 4eV,so 12/3=4, and there will be no current flow? Is this correct? Please help!Thank you.
If the wavelength goes up, the frequency and therefore energy goes down by a factor of three. If that frequency is now below the threshold frequency required to free an outer electron from the metal, there will be no emission at all and no current no matter what intensity or stopping voltage.
Now those electrons had a Ke of 8 eV after being freed from the metal.
So 4 ev of the 12 was used just to spring the electrons out of their orbits.
If you only hit the metal with 4 eV photons, you just barely release the outer electrons and a tiny stopping voltage approaching zero will reduce the current to zero.
THANK YOU DAMON!
To understand what would happen if the wavelength of the incident photons was tripled, we need to consider the relationship between the energy and wavelength of photons.
The energy (E) of a photon is given by the equation E = hc/λ, where h is Planck's constant (approximately 6.626 x 10^-34 J·s), c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength of the photon.
In this case, the incident photons have an energy of 12 eV, which is equivalent to 12 x 1.6 x10^-19 Joules (since 1 eV = 1.6 x 10^-19 J).
If the wavelength is tripled, it means λ_new = 3λ_old. Let's now analyze the situation step by step:
1. Determine the energy of the photons if the wavelength is tripled:
E_new = hc/λ_new
= hc/(3λ_old)
This implies that E_new is one-third of E_old:
E_new = (1/3)(hc/λ_old)
= (1/3)E_old
= (1/3)(12 x 1.6 x10^-19 J)
= 4 x 1.6 x10^-19 J
= 6.4 x10^-19 J
So the energy of the photons becomes 6.4 x10^-19 J, not 4 eV.
2. Compare the new energy with the work function of the metal:
The work function (ϕ) of a metal is the minimum energy required to remove an electron from its surface. If the energy of the photons is greater than or equal to the work function of the metal (E_new ≥ ϕ), current will flow.
In this case, if the stopping potential is 8.0V, it means that the work function of the metal is also 8.0 eV.
Since the energy of the photons (6.4 x10^-19 J) is less than the work function (8.0 eV), the photons do not have sufficient energy to remove electrons from the metal's surface. As a result, there will be no current flow.
Therefore, your intuition that there will be no current flow is correct. However, the energy of the tripled-wavelength photons would be 6.4 x10^-19 J, not 4 eV.
I hope this helps! Let me know if you have any further questions.