Help with 33 (6th edition)  [ENDORSED]

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KylieY_3B
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

Help with 33 (6th edition)

Postby KylieY_3B » Sun Oct 21, 2018 12:20 pm

Could someone explain how to solve HW problem 33? In particular part A is causing me trouble.

Chemistry Community won't let me copy and paste the text of a question but it essentially asks to find the wavelength of an emitted electron with a velocity of 3600 kilometers per second.

Mahir_Hasan2C
Posts: 60
Joined: Fri Sep 28, 2018 12:25 am

Re: Help with 33 (6th edition)  [ENDORSED]

Postby Mahir_Hasan2C » Sun Oct 21, 2018 12:55 pm

The first step is to convert it to m/s. Then you would use the equation lambda (wavelength) = h/(mass of the electron*velocity). This would give your answer.

305144105
Posts: 32
Joined: Thu May 10, 2018 3:00 am

Re: Help with 33 (6th edition)

Postby 305144105 » Mon Oct 22, 2018 6:07 pm

1. Convert 3.6*10^3 km s^-1 to m s^-1
3.6*10^3 km s^-1 = 3.6*10^6 m s^-1

2. Use this value in the formula λ = h/mv. This formula is derived from given formulas λ = h/p and p=mv.
λ = h/mv = (6.626 * 10^-34) / ((9.109*10^31) * (3.6 * 10^6)) = 2.02 * 10^-10m

KylieY_3B
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

Re: Help with 33 (6th edition)

Postby KylieY_3B » Mon Oct 22, 2018 6:47 pm

Thank you!

For part B (determining how much energy needed to remove electron from metal surface), would I use the equation Ek = .5 (Me)(Ve)^2?


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