The photogenic work function of a metal is the minimum energy needed to eject an electron by irradiating the metal with light. For calcium, this work function equals 4.34 x 10^-19 J. What is the minimum frequency of light for the photoelectric effect in calcium?
I'm confused so if you could break it down how you got the solution I will greatly appreciate it. Thanks again.
To find the minimum frequency of light for the photoelectric effect in calcium, we can use the equation:
E = hf
where E is the energy of a photon, h is Planck's constant (6.63 x 10^-34 J∙s), and f is the frequency of the light.
In the photoelectric effect, the energy of a photon (E) must be equal to or greater than the work function (W) of the metal for an electron to be ejected:
E ≥ W
Given that the work function (W) for calcium is 4.34 x 10^-19 J, we can set up the equation:
E ≥ 4.34 x 10^-19 J
Substituting the first equation into the second equation, we have:
hf ≥ 4.34 x 10^-19 J
Now, we need to solve for the minimum frequency (f). To do this, we can rearrange the equation:
f ≥ (4.34 x 10^-19 J) / h
Plugging in the values for Planck's constant and the work function, we get:
f ≥ (4.34 x 10^-19 J) / (6.63 x 10^-34 J∙s)
f ≥ 6.56 x 10^14 Hz
Therefore, the minimum frequency of light for the photoelectric effect in calcium is approximately 6.56 x 10^14 Hz.
To find the minimum frequency of light for the photoelectric effect in calcium, we can use the equation that relates the energy (E) of a photon to its frequency (ν) as follows:
E = h * ν
Where:
- E is the energy of the photon
- h is Planck's constant (6.626 x 10^-34 J·s)
- ν is the frequency of the photon
The work function of calcium (W) is the minimum energy needed to eject an electron, so we can equate it to the energy of a photon using the above equation:
W = h * ν
Rearranging the equation to solve for ν, we get:
ν = W / h
Substituting the given value for the work function of calcium (W = 4.34 x 10^-19 J) and Planck's constant (h = 6.626 x 10^-34 J·s), we can calculate the minimum frequency:
ν = (4.34 x 10^-19 J) / (6.626 x 10^-34 J·s)
Calculating this expression, we get:
ν ≈ 6.556 x 10^14 Hz
So, the minimum frequency of light for the photoelectric effect in calcium is approximately 6.556 x 10^14 Hz.
The correct word is photoelectric, not photogenic. You used the correct word later in the question. Photogenic means something else entirely.
To solve this problem, set the photon energy requirement equal to the work function, 4.34 x 10^-19 J.
Then set that equal to the photon energy given by the Plank equation
h*f = 4.34 x 10^-19 J
where h is Planck's constant,
h = 6.62*10^-34 J*second
f = 4.34 x 10^-19 J/6.62*10^-34 J*second
The answer will be in Hz (cycles per second)