electron acceleration

Watch this thread
xlaser31
Badges: 6
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 3 months ago
#1
It's a one mark MC question from OCR Oct 2020.
Can't figure out how the correct answer is apparently 1.1x10^7 m/s

An electron with initial kinetic energy of 100 eV and initial speed of 5.9 × 10^6 m/s is accelerated through a potential difference of 250 V. What is the final speed of this electron?
Of course eV=1/2mv^2 and 100eV does result in 5.9 × 10^6 m/s so they seem to have given excess information.
For 1 mark what am I missing?
The difference from the usual scenario is that the electron is not stationary.
0
reply
Pangol
Badges: 15
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 3 months ago
#2
(Original post by xlaser31)
It's a one mark MC question from OCR Oct 2020.
Can't figure out how the correct answer is apparently 1.1x10^7 m/s

An electron with initial kinetic energy of 100 eV and initial speed of 5.9 × 10^6 m/s is accelerated through a potential difference of 250 V. What is the final speed of this electron?
Of course eV=1/2mv^2 and 100eV does result in 5.9 × 10^6 m/s so they seem to have given excess information.
For 1 mark what am I missing?
The difference from the usual scenario is that the electron is not stationary.
Yeah, I'm not sure how the initial speed is helpful. It's easy to work out the new KE of the electron in eV, and then to find out its speed. All assuming that the acceleration is in the same direction as its initial motion. Is there some quick proportionality argument I'm missing?
0
reply
MouldyVinegar
Badges: 7
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 3 months ago
#3
(Original post by xlaser31)
It's a one mark MC question from OCR Oct 2020.
Can't figure out how the correct answer is apparently 1.1x10^7 m/s

An electron with initial kinetic energy of 100 eV and initial speed of 5.9 × 10^6 m/s is accelerated through a potential difference of 250 V. What is the final speed of this electron?
Of course eV=1/2mv^2 and 100eV does result in 5.9 × 10^6 m/s so they seem to have given excess information.
For 1 mark what am I missing?
The difference from the usual scenario is that the electron is not stationary.
As Pangol hinted at, you would use a proportionality argument.

Velocity squared is proportional to the kinetic energy of the particle.

Kinetic energy increases to 350eV. Therefore,  \frac{v^{2}_{final}}{v^{2}_{inital}} = \frac{350eV}{100eV}

Rearranging,  v_{final} = \sqrt \frac{350}{100} \times v_{initial} and just substitute the values into your calculator to find it.
Last edited by MouldyVinegar; 3 months ago
1
reply
Kyaurix
Badges: 4
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 3 months ago
#4
why does the kinetic energy increase to 350 bro?
0
reply
Driving_Mad
Badges: 17
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 3 months ago
#5
(Original post by Kyaurix)
why does the kinetic energy increase to 350 bro?
E = QV, when you sub in you get E = 250 which is the change in kinetic energy, so final kinetic energy = 250 + 100 = 350.

GGG
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

Y13s: How will you be receiving your A-level results?

In person (73)
67.59%
In the post (5)
4.63%
Text (15)
13.89%
Something else (tell us in the thread) (15)
13.89%

Watched Threads

View All