Question

http://prntscr.com/d0pcw5
Any equations or concepts I should be aware of?

Answer 1

@kittiwitti1
Use the relationship between kinetic energy and velocity:
\[\Large KE_e=\frac{ 1 }{ 2 } m_e v^2\]
Here is \(m_e\) the mass of the electron.

Answer 2

I think you need to know (and include) the work function of Hg.

\(\sf \large E_{photon}=KE_e+\phi\)

\(\sf \large \dfrac{hc}{\lambda }=\dfrac{1}{2}m_ev^2+\phi\)

Answer 3

Oh, alright. I will try those...
@aaronq Hmm, I am not sure what this symbol means in your equations... \(\large{\phi}\)?

Answer 4

@kittiwitti1 \(\phi\) is the photoelectric work function, which is the minimum photon energy required to liberate an electron from a substance, it is kinda analogous to a threshold frequency multiplied by the Plank constant.

But since you already have the KE of the electron it should be sufficient to calculate the velocity in a Newtonian mechanical manner.

Answer 5

Ok. I am not sure why you are tagging me in my own post though. @Frostbite

Answer 6

Oh right, i didn't see that K here was kinetic energy.. my bad lol

Answer 7

Sorry kitti. It is a habit I have adopted as I have a similar tagging system in the research group I am working in and always specify to whom I am replying to.

Answer 8

That is ok if there is a legitimate reason lol. I will accept that reasoning :-D @Frostbite haha

Answer 9

On the topic of the question, though, I seem to still be getting the wrong answer. :-( http://prntscr.com/d1su3o

Answer 10

@Frostbite ?

Answer 11

Let me fast do a calculation :)

Answer 12

👌

Answer 13

Are you sure you calculated correctly?
I get the velocity to approx 856335 m/s just by fast calculation

Answer 14

Not sure about my own number either, but yeah...

Answer 15

Here is my work: \[KE=\frac{1}{2}m_{e}v^{2}\]\[3.34\times10^{-19}=\frac{1}{2}(9.109383\times10^{-28})v^{2}\]

Answer 16

\[v=\sqrt{\frac{2(3.34\times10^{-19})}{9.19383\times10^{-28}}}\approx27079.7m/s\]

Answer 17

I am wondering if I got \(\large{m_{e}}\), mass of electron, incorrect...

Answer 18

I think so. 9.10938356 × 10-31 kilograms according to le wiki

Answer 19

Unless... ahhh

Answer 20

Wrong unit I think :P

Answer 21

I thought so too, since my KE is Joules per photon :-(

Answer 22

add units to the velocity and use non-SI units on the energy.
The mass you got is in gram, but joule is kg*m^2 / (s^2)

Answer 23

Those units always grab you with your pants down <.< challenger explosion ftw!

Answer 24

LOL

Answer 25

Ok so the formula is still the same? I am confused

Answer 26

\[KE_{e}=\frac{1}{2}m_{e}v^{2}\]Still applicable?

Answer 27

the SI units for energy is joule. But that is the same as:
\[ \Large 1~ J=1~\frac{ kg~ m^2 }{ s^2 }\]

So in order for the units to cancel to give out a velocity we need to use the mass of the electron in kg. I noticed the number you used is \(10^3\) larger so it must be in gram, while the number I used is in kg.

The equation still holds, simply convert the mass of the electron to kilogram instead of gram and try again.

Answer 28

So... KE will now be this?\[3.34\times10^{-19}\frac{\frac{kgm^{2}}{s^{2}}}{photon}\]

Answer 29

Excuse the small print lol

Answer 30

Yeah. pretty much. You can however drop the per photon unit as we are working on 1:1 relationship between electron and photon.

The thing is, always remember the stupid fact kilogram is the SI units, so 1 joule will also be in kilogram therefore the mass of the electron had to be written in kg.

Answer 31

So basically I need to convert the unit form of the \(\large{m_{e}}\) and I'm good?

Answer 32

Yup :D

Answer 33

Whew! I'm glad it was a simple error x.x I still have several problems to go over AND a quiz on this material today, so...

Answer 34

Oh, and a lab that's ALSO due before class. EVERYTHING is due before class. Ew.

Answer 35

hehe the error is always in the small things. Also what I kinda expected, since you are in fact really good at chemistry. ;)

Answer 36

I just hope the textbox turn green now....

Answer 37

lol naw, I just applied the chemistry formulae that were given to me. I'm not smart at this at all :-P

Answer 38

So then, this?\[3.34\times10^{-19}=\frac{1}{2}(9.109383\times10^{-28}g\times\frac{10^{-3}kg}{1g})v^{2}\]

Answer 39

I hope I got the conversion unit correct.

Answer 40

Looks good

Answer 41

Yay! :-D

Answer 42

\[v=\sqrt{\frac{2(3.34\times10^{-19})}{9.109383\times10^{-31}}}\]Looks good?

Answer 43

It is very good.

Answer 44

Okay :-D
I will put it in Wolfram Alpha now, brb lol

Answer 45

Let me know if it was indeed the correct answer. :D

Answer 46

http://www.wolframalpha.com/input/?i=sqrt%5B(2*3.34e-19)%2F(9.109383e-31)%5D \[8.56335\times10^{5}\]

Answer 47

Sounds much like something I got.

Answer 48

YES!! It is correct (/ > 0 <)/

Answer 49

LOL, Bill Nye.

Answer 50

THANK YOU!! @Frostbite (/ *-*)/
I have more questions though ahah...

Answer 51

I'll make a new post. :-)

Answer 52

@kittiwitti1 no problem at all... I will be working a little on my protein work until I am ones required :)

Answer 53

Okay, it will be coming soon :-P lol