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?
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 use the concept of energy conservation.
The energy of a photon is given by the equation: E = hf, where E is the energy of the photon, h is Planck's constant (6.63×10^(-34) J⋅s), and f is the frequency of the light.
Given that the energy of the light burst is 5.02× 10–7 J, we can rearrange the equation to solve for the frequency:
f = E / h
f = 5.02× 10–7 J / (6.63×10^(-34) J⋅s)
f ≈ 7.57×10^26 s^(-1)
Now, to find the maximum number of electrons that could be ejected, we need to compare the energy of the light burst (5.02× 10–7 J) with the energy required to eject a single electron.
The energy required to eject an electron is given by the equation: E_ejected = hf_ejected, where E_ejected is the energy required to eject an electron and f_ejected is the frequency of the light at which the electrons were emitted.
Given that f_ejected is 4.26× 10^15 s^(-1) and the kinetic energy of the ejected electrons is 4.50× 10^(-19) J, we can rearrange the equation to solve for the energy required to eject an electron:
E_ejected = hf_ejected
E_ejected = (6.63×10^(-34) J⋅s) * (4.26× 10^15 s^(-1))
E_ejected ≈ 2.82× 10^(-18) J
Now we can calculate the maximum number of electrons that could be ejected:
Number of electrons = Energy of light burst / Energy required to eject an electron
Number of electrons = 5.02× 10^(-7) J / 2.82× 10^(-18) J
Number of electrons ≈ 1.78× 10^(11)
Therefore, the maximum number of electrons that could be ejected from this metal by a burst of light with a total energy of 5.02× 10^(-7) J is approximately 1.78× 10^(11).