## Post-Mod Problem #28

Summer de Vera 2C
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

### Post-Mod Problem #28

Hey! I'm having a bit of trouble deciding which equation(s) to use in order to figure out this problem from the post-module assessment. It seems like I have to do this in several steps, however I'm not sure where to start.

28. Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.
A. What is the kinetic energy of the ejected electron?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

### Re: Post-Mod Problem #28

You know that E(photon) - E(energy to remove e-) = E(kinetic energy of e-). You also know that E(photon) = h(planck's constant) x v (frequency)
In other words, hv - work function = (1/2)mv^2
(where m = mass of e- and v = velocity of e-)
Mass of e- = 9.11E-31, Given velocity = 6.61E5
This would then lead to E(kinetic energy of e-) = (1/2)mv^2
Plug in the values to get (1/2)(9.11E-31)(6.61E5)^2 = 1.99E-19 J.

Alexa_Henrie_1I
Posts: 61
Joined: Fri Sep 29, 2017 7:03 am

### Re: Post-Mod Problem #28

and mass of e- will always be 9.11x10^-31 for that equation right?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

### Re: Post-Mod Problem #28

The mass of an electron is always about 9.11E-31 kg.

005115864
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

### Re: Post-Mod Problem #28

Hi! I'm still confused about how that got us the answer. I understand the equations we're supposed to use and where everything goes but the one part I am still confused about is the work function.

If the work function is 150.6 KJ, shouldn't we first convert that to joules which would be 1.506x10^-5.

Then to my understanding aren't we supposed to use

total energy (c=lamda*v) - work energy = Ek (.5MV^2) ?

If that's the case, wouldn't we add the work energy to Ek for the answer?

I am still new to this so I am simply trying to gain a firmer grasp on the manipulation of equations.