When a metal was exposed to light at a frequency of 4.26× 1015 s–1, electrons were emitted with a kinetic energy of 4.50× 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 5.02× 10–7 J?
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To find the maximum number of electrons that could be ejected from the metal by a burst of light with a total energy of 5.02×10^(-7) J, we need to calculate the energy required to eject a single electron and then divide the total energy by it.
First, we find the energy required to eject a single electron by using the kinetic energy of the electrons emitted when the metal was exposed to light at a frequency of 4.26× 10^15 s^(-1).
The energy of a single electron can be calculated using the formula:
E = hf - φ
Where:
E is the energy of a single electron,
h is Planck's constant (6.626 × 10^(-34) J·s),
f is the frequency of light (4.26 × 10^15 s^(-1)),
φ is the work function (energy required to remove an electron from the metal).
Rearranging the formula, we have:
φ = hf - E
Plugging in the known values:
φ = (6.626 × 10^(-34) J·s) × (4.26 × 10^15 s^(-1)) - (4.50 × 10^(-19) J)
φ = 2.82 × 10^(-18) J
Now that we know the work function, we can determine the number of electrons that can be ejected from the metal.
Number of electrons = Total energy / Energy of a single electron
Number of electrons = (5.02×10^(-7) J) / (2.82 × 10^(-18) J)
Calculating this, we get:
Number of electrons ≈ 1.78 × 10^(11)
Therefore, the maximum number of electrons that could be ejected from the metal by the burst of light with a total energy of 5.02×10^(-7) J is approximately 1.78 × 10^(11).