Test 2, Question #7  [ENDORSED]


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Hena Sihota 1L
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Test 2, Question #7

Postby Hena Sihota 1L » Sat Oct 28, 2017 11:40 pm

Problem #7:
A) An electron was ejected from the surface of an unknown metal. The ejected electron had 2.35 x 10^(-18) J of kinetic energy. The photon used to eject the electron had 4.19 x 10^(-18) J of energy. Calculate the work function (or threshold energy) of the unknown metal.
B) What is the velocity of the ejected electron in part a?
I got part "A" correct with the answer 1.84 x 10^(-18) J, but I got part "B" wrong. I used the equation E=hv to find the wavelength of the ejected electron so I could use the equation V=(h)/(mlamda) to solve for the velocity; however, my TA commented "E=hv only applies to massless particles." Can someone please explain how to correctly solve part "B"?

Nishma Chakraborty 1J
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Re: Test 2, Question #7  [ENDORSED]

Postby Nishma Chakraborty 1J » Sat Oct 28, 2017 11:57 pm

Hey, so for part b, I didn’t use the de Broglie equation. Carrying on from part a. (which referred to electron spectroscopy and the equation: (Energy of the Photon - Work function = Kinetic Energy), I used the definition of kinetic energy (1/2m*v^2) in order to find the velocity of the ejected electron. I had:

1/2*(mass of the electron)*(velocity of the electron)^2 = KE
KE was given as 4.19*10^18 J in the problem.
Mass of an electron is 9.11*10^-31 kg.
So, (4.0*10^18)=1/2(9.11*10^-31)*(velocity of the electron)^2
By solving for v, I got 3.03*10^6 m/s.

Hope this helps!

Justin Bui 2L
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Joined: Fri Sep 29, 2017 7:06 am

Re: Test 2, Question #7

Postby Justin Bui 2L » Sun Oct 29, 2017 10:15 pm

Yup, I did the same thing as ^^ and I think it's because 1/2mv^2 describes best the energy of the moving electron while DeBroglie's equation also uses velocity but it's moreso for waves/light

nelms6678
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Re: Test 2, Question #7

Postby nelms6678 » Sun Oct 29, 2017 10:26 pm

Test 2, Question #5
Determine the frequency of the emitted photon when an electron in a hydrogen atom drops from the quantum state of n=3 to n=1.

I used the formula -hR/n^2 and calculated:
n=3... -1/9hR = -2.42x10^-19
n=1... -1hR = -2.18x10^-18
Using these two values I then used N(final)-N(intial), so -2.18x10^-18 - (-2.42x10^-19)
= -1.94x10^-18... I got the wrong value but my steps seem to be correct where did I make a mistake?

nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Re: Test 2, Question #7

Postby nelms6678 » Sun Oct 29, 2017 10:40 pm

Regarding the work posted above for part b of question #7-
So did you take (4.0*10^18) divided by 1/2(9.11*10^-31), then once you get that value square root it because velocity is V^2? I tried that and was unable to get 3.03*10^6

Nishma Chakraborty 1J
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

Re: Test 2, Question #7

Postby Nishma Chakraborty 1J » Tue Oct 31, 2017 8:01 pm

Hey, so I think my test asked about an energy emitted as an electron moved from n=4 to n=1, but I solved for the version you had - did you make sure to calculate for frequency? I got -1.94*10^18 J for Energy (the negative due to the fact that the electron is emitting energy) as well. You set 1.94*10^18 J (Energy) equal to hv, so you divide the calculated value by Planck's constant in order to get frequency. When I did that, I got 2.924*10^15 Hz.

For #7,
(4.19*10^-18)=1/2(9.11*10^-31)*(velocity of the electron)^2
I multiplied both sides by 2 in order to get rid of the 1/2 on the right side of the equation, so:
(8.38*10*-18)=(9.11*10^-31)*v^2
then divided by 9.11*10^-31 to get:
9.199*10^12=v^2,
followed by taking the square root of both sides, getting:
v=3.03*10^6 m/s

Hope this helps! :)


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