Howard Wolowitz is developing a radiation detector for the Intemational space Station. He uses a thin film composed of a metal alloy that has a threshold frequency of 5.5x10^15 Hz. When radiation is shone on the thin film, the fastest electron emitted has a velocity of 668000 m/s.
A. What is the KE, in J, of the fastest ejected electron?
I was able to figure this one out which was 2.03x10^-19
B. What is the wavelength, in nm , of the photon that caused the ejection of this electron?
I tried using wavelength= h/mv. But I didn't get the right answer. The right answer for this is 52nm. Which I do not know how to get!
C. What is the wavelength, in nm, of this electron?
The right answer for this one is 1.09 nm
I just need guidance for answering these question. If you are able to help thank you!
A. ok I got 2.09*10^-19 J but used crude electron mass approx
B. Ephoton = Eelectron + h * 5.5*10^15
Ephoton = 2.03*10^-19 + 6.63*10^-34*5.5*10^15
Ephoton = 2.03*10^-19 + 36.5*10^-19
Ephoton = 38.5 *10^-19 Joules
now what wavelength is that
E = h c/lambda so lambda = h c/E
lambda = 6.63*10^-34 * 3*10^8 / 38.5*10^-19
= .516*10^-7 = 51.6 *10^-9 meters
= 51.6 nanometers so I agree with them
c. lambdfa = h/mv = 1.09 nm
b. The threshold is 5.5E15 Hz
c = freq x wavelength gives me about 54 nm.
for electron
lambda = h/mv = 6.63*10^-34 /[ 9.1*10^-31*6.68*10^5]
= .109*10^-8 = 1.09 * 10-9 = 1.09 nm
A. To find the kinetic energy (KE) of the fastest ejected electron, you can use the equation KE = (1/2)mv^2, where m is the mass of the electron and v is its velocity. The mass of an electron is approximately 9.11 x 10^-31 kg.
So, KE = (1/2)(9.11 x 10^-31 kg)(668000 m/s)^2 = 2.03 x 10^-19 J (which you correctly found).
B. To determine the wavelength (λ) of the photon that caused the ejection of the electron, you can use the equation λ = c/f, where c is the speed of light (approximately 3 x 10^8 m/s) and f is the frequency of the photon. In this case, the threshold frequency is given as 5.5 x 10^15 Hz.
So, λ = (3 x 10^8 m/s) / (5.5 x 10^15 Hz) = 54.5 x 10^-7 m = 54.5 nm (approximately). So the correct answer is 55 nm, not 52 nm as you mentioned. Different sources may give slightly different values due to rounding or approximation.
C. To find the wavelength of the electron, you can use the de Broglie wavelength equation: λ = h/p, where h is the Planck's constant (approximately 6.626 x 10^-34 J·s) and p is the momentum of the electron. The momentum of an electron can be calculated as p = mv.
So, λ = (6.626 x 10^-34 J·s) / [(9.11 x 10^-31 kg)(668000 m/s)] = 1.09 x 10^-9 m = 1.09 nm (approximately). Thus, the correct answer is indeed 1.09 nm.
If you have any further questions, feel free to ask!
Sure, I can help you with these questions. Let's go through each one step by step.
A. To find the kinetic energy (KE) of the fastest ejected electron, we can use the equation:
KE = 1/2 * m * v^2
In this case, we are given the velocity (v) of the electron, which is 668,000 m/s. We also know the mass (m) of an electron, which is approximately 9.11 x 10^-31 kg.
Plugging in these values into the equation, we get:
KE = 1/2 * (9.11 x 10^-31 kg) * (668,000 m/s)^2
Simplifying and calculating this expression, we find the answer to be approximately 2.03 x 10^-19 J, as you correctly stated.
B. To find the wavelength (λ) of the photon that caused the ejection of this electron, we can use the equation:
λ = c / f
Where c is the speed of light in a vacuum, which is approximately 3 x 10^8 m/s, and f is the frequency of the photon.
We are given the threshold frequency of the alloy, which is 5.5 x 10^15 Hz. The frequency of the photon that caused the ejection of the electron would be the same as the threshold frequency, since it just barely had enough energy to eject an electron.
Plugging in these values into the equation, we get:
λ = (3 x 10^8 m/s) / (5.5 x 10^15 Hz)
Simplifying and calculating this expression, we find the answer to be approximately 5.45 x 10^-7 m, which can be converted to 545 nm (since 1 nm = 10^-9 m), not 52 nm. Could you please double-check the answer you have?
C. To find the wavelength of the electron, we can make use of the de Broglie wavelength equation:
λ = h / p
Where λ is the wavelength, h is Planck's constant (approximately 6.626 x 10^-34 J·s), and p is the momentum of the electron.
The momentum (p) can be calculated using the mass (m) and velocity (v) of the electron:
p = m * v
Given that we know the mass of an electron (9.11 x 10^-31 kg) and the velocity of the electron (668,000 m/s), we can find the momentum. Then, using the momentum in the de Broglie wavelength equation, we can calculate the wavelength of the electron.
I hope this helps! Let me know if you have any further questions or need any more clarification.