When a metal was exposed to light at a frequency of 3.86× 1015 s–1, electrons were emitted with a kinetic energy of 3.60× 10–19 J. What is the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 4.88× 10–7 J?
To find the maximum number of electrons that could be ejected from the metal, we need to first find the energy required to eject a single electron.
We can use the equation:
E = hf - Φ
Where:
E = Energy of the ejected electron (3.60× 10–19 J)
h = Planck's constant (6.626 × 10^-34 J·s)
f = Frequency of the light (3.86× 1015 s–1)
Φ = Work function of the metal (unknown)
Rearranging the equation to solve for the work function:
Φ = hf - E
Φ = (6.626 × 10^-34 J·s) * (3.86× 1015 s–1) - (3.60× 10–19 J)
Φ ≈ 2.55 × 10^-19 J
Now we have the work function of the metal, which is the minimum energy required to eject an electron.
To find the maximum number of electrons that could be ejected by the burst of light with a total energy of 4.88 × 10^-7 J, we divide the total energy by the work function:
Number of electrons = Total Energy / Work Function
Number of electrons = (4.88 × 10^-7 J) / (2.55 × 10^-19 J)
Number of electrons ≈ 1.91 × 10^12
Therefore, the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 4.88 × 10^-7 J is approximately 1.91 × 10^12.
To solve this problem, we need to make use of the concept of the photoelectric effect and utilize the equation:
E = hv - φ
where:
- E is the energy of the incoming photon (light burst)
- h is the Planck's constant (6.626 × 10^-34 J·s)
- v is the frequency of the incoming photon
- φ is the work function or the minimum energy required to remove an electron from the metal's surface
In this problem, we are given:
- The kinetic energy of the emitted electrons (3.60 × 10^-19 J)
- The frequency of the light that caused the emission (3.86 × 10^15 s^-1)
First, we need to determine the work function:
E = hv - φ
φ = hv - E
Plugging in the values:
φ = (6.626 × 10^-34 J·s)(3.86 × 10^15 s^-1) - (3.60 × 10^-19 J)
φ ≈ 2.555 × 10^-15 J
Now that we know the work function, we can calculate the maximum number of electrons that could be ejected from the metal by the burst of light with a total energy of 4.88 × 10^-7 J.
To find this, we divide the total energy by the energy required to remove a single electron:
N = E_total / φ
Plugging in the values:
N = (4.88 × 10^-7 J) / (2.555 × 10^-15 J)
N ≈ 1.91 × 10^8
Therefore, the maximum number of electrons that could be ejected from this metal by the light burst is approximately 1.91 × 10^8.