Determine the shortest frequency of light required to remove an electron from a sample of Ti metal, if the binding energy of titanium is 3.14 × 103 kJ/mol

I see this question answered on Google, but it looks like an incredible eyesore to look at and I can't understand the work behind it.

Halp plz?

This post was edited by desertwolf on Nov 23 2013 10:51am

was the 3.14 supposed to have an exponent ? 3.14 x 103= 323.42 kJ/mol to break off an electron. so we need to find the frequency of light that is powerful enough to do this.

we do not want to break off an entire molecule of Ti, only an electron. so convert to joules and divide energy/avogadros number

323.42 kJ = 323420 J / 6.022 x 10^23 = 5.37 x 10 ^-19 J to break off a single Ti electron

next step is to plug in the energy required for one electron into our formula:

E=hv

5.37 x 10 ^ -19 J = (6.626x10^-3 j*s)v
divivde by plancks constant on both sides to isolate v
v= 8.1044x10^-17 s^-1

now we have frequency. We can now plug this into our c=l (lambda) v equation

speed of light is never unknown because we know the value. it is 3x 10 ^8 m/s

so, 3x10^8 m/s = l(lambda) (8.1044x10^-17 s^-1)
divide by by V to isolate lambda (lambda is the wavelength)

we end up with lambda = 3.702x 10^24 m x 1x10^-9 nm/ 1m = 3.702x 10^15 nm






ah i just realized it was 3.14 x 10^3

written 10^3 not 103. not going to go back and redo the problem but you can hopefully see now how it works

This post was edited by fingerling on Nov 23 2013 02:42pm

actuaally looks like it will end up just switching the exponent

3140 kJ= 3140000 J / 6.022 x 10^23 = 5.214x 10^-18

5.214x 10^-18 / 6.626x10^-3= 7.869x10^-16

c=lv

3.812x10^23 m = lambda = 1x10^-9/ 1 m = 3.812x10^14 nm

Something doesn't look right. Its been a while since I've done this but iirc a very small wavelength is going to have a much higher energy. So we would expect the answer to be something with a negative exponent. I'll ask my prof on Monday.


Using phone calc with no dedicated negative key only minus. May have fkd results. The steps I used are correct though so maybe go back through and do the arithmetic on your calc

This post was edited by fingerling on Nov 23 2013 02:51pm

Ya my apologies. It is 10^3. Copy/paste failed on me. I'll look over the work and see if I can understand this.


The answer is7.87 x 10^15 Hz btw.

This post was edited by desertwolf on Nov 23 2013 03:02pm

I thought plank's constant was 6.626 x 10^-34. Maybe this is where you screwed up?

Looks like it he he sorry .


If you do it with the correct value you get 7.87x10^15.

So yeah the work is correct just had the wrong value. Also I went too far as I read the question wrong.


3.811 x 10^ -8 m x 1x10-9 /1m = 3.811x 10^-17 = wavelength

OK this looks much better this wavelength would be damaging


Goodluck



This post was edited by fingerling on Nov 23 2013 03:26pm

Alright, so...from what I'm seeing...

1)
First you converted 3.14 × 10^3 kJ/mol to 3140000 J/mol.

2)
Divide 3140000 J/mol by Avogrado's number (6.022 x 10^23) to get Joules by and the number 5.214 x 10^-18

3)
Manipulate the E = h v

Plugging in 5.214 x 10^-18J into E and solve for frequency. Meaning I just divide E by planks number #(6.626 x 10^-34). Making me get 7.87x10^15.

amirite?

Ok, I understand the math, but how did you understand the word problem the first place? This thing is mentioning Titanium and shit =/.

The substance given is arbitrary. You just need a value to plug in for E. Any value will do. Ignore the specific type of element or w/e they give you and focus on the value for e. A lot of chemistry is just thinking about what you are trying to get to with the given info. If we are given an energy to break up a ti molecule, but they ask about breaking up a ti electron, we know we must divide the energy by avogadros number to isolate the energy for breaking up one electron as opposed to one whole atom. The rest is just plugging in numbers, like you listed in the steps. Sometimes they ask about wavelength, in which case you would do exactly the same procedure but would then use the 7.87 frequency value to plug into c=lambda V equation