The minimum frequency for the photoelectric effect of a zinc plate is 9.7 × 1014 Hz. What will be the kinetic energy of an ejected electron when a photon with a wavelength 200 nm (2.00 × 10 7 m) strikes it?
200 nm = 2• 10 ^-8 m (! ! !)
Einstein's photoelectric equation:
ε =W + KE,
ε = h•c/λ = 6.63•10^-34•3•10^8/2•10^-7 =9.9•10^-19 J.
W =h•f= 6.63•10^-34•9.7•10^14 =6.43•10^-19 J.
KE = ε – W = 9.9•10^-19 - 6.43•10^-19 =3.47•10^-19 J.
Well, well, well, looks like we've got ourselves a physics question here! Let's see if I can inject some humor into the photoelectric effect.
Alrighty then, to find the kinetic energy of an ejected electron, we need to use the equation E = hf - φ, where E is the kinetic energy, h is Planck's constant, f is the frequency of the incident photon, and φ is the work function of the metal. Since we're dealing with zinc, we know the minimum frequency is 9.7 × 10^14 Hz.
Now, to find the frequency of the photon, we can use the formula c = fλ, where c is the speed of light and λ is the wavelength of the photon. Rearranging that equation, we find that f = c / λ.
Plug in the values for c (the speed of light) and λ (the wavelength), and you'll get the frequency. Then, you can calculate the kinetic energy using the given values of h and φ.
Now, I apologize if my humor fell a bit flat here, but sometimes physics just zaps all the jokes out of you. Nevertheless, I hope this helps you calculate the kinetic energy of that ejected electron!
To find the kinetic energy of an ejected electron in the photoelectric effect, we need to use the formula:
Einstein's Photoelectric Equation: E = hf - φ,
where:
E is the kinetic energy of the electron,
h is Planck's constant (6.626 × 10^-34 J·s),
f is the frequency of the incident photon,
and φ is the work function (the minimum energy required to remove an electron from an atom).
We are given the frequency of the photoelectric effect for a zinc plate, which is 9.7 × 10^14 Hz. However, we need to find the frequency corresponding to a given wavelength of 200 nm.
To find the frequency, we can use the equation:
Speed of Light (c) = λf,
where:
c is the speed of light (3.00 × 10^8 m/s),
λ is the wavelength of the incident photon,
and f is the frequency.
Rearranging the equation, we get:
f = c / λ.
Plugging in the values:
f = (3.00 × 10^8 m/s) / (200 nm (2.00 × 10^-7 m)).
Simplifying the expression:
f = 1.50 × 10^15 Hz.
We now have the frequency of the incident photon. Substituting this value into the photoelectric equation, we get:
E = (6.626 × 10^-34 J·s) × (1.50 × 10^15 Hz) - φ.
Unfortunately, the work function (φ) for the zinc plate is not provided. Therefore, we cannot calculate the exact value for the kinetic energy of the ejected electron.
To find the kinetic energy of an ejected electron in the photoelectric effect, we can use the equation:
Kinetic Energy (KE) = Energy of Photon (E) - Work Function (W)
where the Energy of Photon (E) is given by the equation:
E = h * f
where h is the Planck's constant (6.626 × 10^-34 J.s) and f is the frequency of the photon.
First, let's find the frequency of the photon using the given wavelength. The formula relating wavelength (λ) and frequency (f) is:
f = c / λ
where c is the speed of light (3.00 × 10^8 m/s).
Plugging in the values, we have:
f = (3.00 × 10^8 m/s) / (2.00 × 10^-7 m)
f ≈ 1.50 × 10^15 Hz
Since the frequency of the photon is greater than the minimum frequency required for the photoelectric effect, the electron will be ejected.
Now, let's calculate the kinetic energy of the ejected electron. We can use the formula:
KE = E - W
where W is the work function, which represents the minimum energy required to remove an electron from the metal surface. The work function for zinc is not given, so we cannot determine the exact value.
However, if we assume that the work function (W) is zero, then the kinetic energy (KE) will be equal to the energy of the photon (E).
So, substituting the values, we have:
KE = (6.626 × 10^-34 J.s) * (1.50 × 10^15 Hz)
KE ≈ 9.94 × 10^-19 J
Therefore, if the work function is assumed to be zero, the kinetic energy of the ejected electron will be approximately 9.94 × 10^-19 Joules.