When light of a wavelength of 125 nm strikes a certain metal, electrons having a velocity of 8.2 x 10^5 m/s are
emitted from the surface of the metal. What is the threshold frequency of the metal?
The answer is 1.9 x 10^15 s^-1
I'm not sure how to do this. The equations that I tried to use were E=1/2mv^2, E=hf-hf0, E=hc/lamda but it didn't work
KE = (hc/w)-hf
KE = 1/2(m)(v^2) so
1/2(m)(V)^2 = (hc/wavelength) - hf.
You have m in kg, v in m/s, h, c, and wavelength (w in meters). Solve for f.
I get 1.9 x 10^15 Hz.
To find the threshold frequency of the metal, we can make use of the work function concept. The work function is the minimum amount of energy required to remove an electron from the metal surface.
The equation relating energy and frequency is given by E = hf, where E is the energy, h is the Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the light.
The equation relating energy and wavelength is given by E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light.
Since we know both the velocity of the emitted electrons and the wavelength of the light, we can use the equation E = 1/2mv^2 (kinetic energy) to determine the energy of the emitted electrons.
Now, let's go step-by-step to solve this problem:
Step 1: Convert the wavelength to meters.
125 nm = 125 x 10^-9 m
Step 2: Use the equation E = hc/λ and rearrange it to solve for the energy.
E = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (125 x 10^-9 m)
E ≈ 1.590 x 10^-16 J
Step 3: Use the equation E = 1/2mv^2 to calculate the kinetic energy of the emitted electrons. Since we have the velocity and mass is not given, we can assume that the mass of the electrons is negligible in this case. Thus, we can ignore the mass in the equation.
1/2mv^2 = 1/2(8.2 x 10^5 m/s)^2
1/2mv^2 ≈ 3.367 x 10^-12 J
Step 4: The threshold frequency is the value at which the kinetic energy (3.367 x 10^-12 J) is equal to or greater than the energy of the light (1.590 x 10^-16 J). So, we can set them equal to each other.
3.367 x 10^-12 J = hf
Step 5: Rearrange the equation to isolate the frequency.
f = (3.367 x 10^-12 J) / (h)
f ≈ 5.086 x 10^21 Hz
Step 6: Finally, convert the frequency to s^-1.
1 Hz = 1 s^-1, so
5.086 x 10^21 Hz ≈ 5.086 x 10^21 s^-1
Therefore, the threshold frequency of the metal is approximately 5.086 x 10^21 s^-1, which is not the same as the given answer of 1.9 x 10^15 s^-1. Please double-check the original question or check if there was any mistake in the calculations.
To find the threshold frequency of the metal, you can apply the equation that relates the energy of a photon to its wavelength or frequency. This equation is given as:
E = hf
Where E represents the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the light.
In this case, you are given the wavelength of the incident light (125 nm), but first, you need to convert it to meters. Since 1 nm is equal to 1 x 10^-9 m:
Wavelength (λ) = 125 nm = 125 x 10^-9 m = 1.25 x 10^-7 m
Now, let's find the energy of the photon using the equation:
E = hc/λ
Where c is the speed of light (3 x 10^8 m/s).
E = (6.626 x 10^-34 J·s * 3 x 10^8 m/s) / (1.25 x 10^-7 m)
Calculating this expression gives us the energy of the photon.
Now we need to determine the threshold frequency of the metal, which is the minimum frequency required to release the electrons.
Threshold frequency (f0) can be determined by rearranging the equation:
f0 = E / h
Substituting the calculated photon energy into the equation:
Threshold frequency (f0) = Energy / Planck's constant
Now you can calculate the threshold frequency:
f0 = (6.626 x 10^-34 J·s * 3 x 10^8 m/s) / (1.25 x 10^-7 m)
The result should give you the threshold frequency of the metal.