Sapling Homework - electron affinity

[SamanthaTolentino 3D]

The E.Coli bacterium is about 2.2 μm long. Suppose you want to study it using photons of that wavelength or electrons having that de Broglie wavelength.

What is the energy of the photon? What is the energy of the electron?

Im struggling on where to start. Do I assume that the length of the bacteria is the wavelength?

[Carolina 3E]

Yes, the length of the bacterium will be the wavelengths of the photon and electron.
To find the energy of the photon, use E = h(frequency) and c = λ(frequency) and solve for E
The energy of the electron is its kinetic energy, which involves mass and velocity. To find the velocity of e-, use de Broglie's equation: λ = h/p , p = mv. Then use the velocity and mass of e- to calculate its kinetic energy

[Jiapeng Han 1C]

So for calculating energy of photons, you could use the equation E=h*f where frequency of light could be determined via c=f*wavelenth.
For calculating energy of electrons, you could firstly calculate the momentum of electron via p=h/wavelength, and then apply the value of p to the equation p^2/2m=Ek.

[Dylan_Nguyen_2C]

For this question, you need to use the length of bacteria to determine both energies. For the proton, you would use λ=c/v to solve for the frequency v, then plug that value into the E=hv equation to solve for the energy of a single photon. Keep in mind you may need to convert micrometers to meters when substituting in for wavelength(2.2 µm=2.2x10^-6 m). For the electron, you would want to use λ=h/mv, using Planck's constant along with the mass of an electron(9.109x10^-31 kg) and the given wavelength to solve for the velocity of the electron. We can then plug velocity v into the kinetic energy equation E=1/2mv^2, again using mass of an electron to solve for the energy of an electron. Hope this helps

[Abraham De Luna]

yeah it seems like the length of the bacterium would be the same wavelengths of the photon.

[Tiffanny_Carranza_2D]

hello I am having a lot of trouble with this problem for the energy of an electron.

I did:
Ke=1/2(m)(v^2) for m i put in the mass of an electron: 9.109x10^-31

I got v from: v= (3.0x10^8 m/s)/(2.1x10^-6 m)= 1.42x10^14s

Ke= 1/2(9.109x10^-31 kg )((1.42x10^14 s) ^2)

the answer I got was Ke= 9.18x10^-3.

i don't know what I am doing wrong

[Marisa Gaitan 2D]

To find velocity, you use the de broglie equation lamba=h/p, substituting mv for p. You then solve for v, and input this value into the 1/2mv^2 equation.

[Marisa Gaitan 2D]

hello I am having a lot of trouble with this problem for the energy of an electron.

I did:
Ke=1/2(m)(v^2) for m i put in the mass of an electron: 9.109x10^-31

I got v from: v= (3.0x10^8 m/s)/(2.1x10^-6 m)= 1.42x10^14s

Ke= 1/2(9.109x10^-31 kg )((1.42x10^14 s) ^2)

the answer I got was Ke= 9.18x10^-3.

i don't know what I am doing wrong

To find velocity, you use the de broglie equation lamba=h/p, substituting mv for p. You then solve for v, and input this value into the 1/2mv^2 equation.

[Tiffanny_Carranza_2D]

hello I am having a lot of trouble with this problem for the energy of an electron.

I did:
Ke=1/2(m)(v^2) for m i put in the mass of an electron: 9.109x10^-31

I got v from: v= (3.0x10^8 m/s)/(2.1x10^-6 m)= 1.42x10^14s

Ke= 1/2(9.109x10^-31 kg )((1.42x10^14 s) ^2)

the answer I got was Ke= 9.18x10^-3.

i don't know what I am doing wrong

To find velocity, you use the de broglie equation lamba=h/p, substituting mv for p. You then solve for v, and input this value into the 1/2mv^2 equation.

so my velocity is wrong?

[Kendall_Dewey_2D]

These explanations helped so much - I was stuck on this one for a while. Thanks guys!