78 Replies
ayo
bot is down??
@iTeachChem Helper
bruhhh
@Moderator please ping chem
@Chemistry Quiz
Arrey
Wrong ping
Ahmm dexter
It's Dexter
Dexter
@Dexter
Can't believe this is why I end up coming back to doubts section
Damn they mixed photoelectric effect and electrochemistry
Feisty
Idea should be
E nikaalo
Ratio lo
This would contain the pH term
You know the other side of the ratio as the work function
I'll do the math......
Although I don't want to buy I guess I'll do it
@SirLancelotDuLac aa jaayo pls if free š
Verify kar dena math
lol
ye work function kuch suna suna lag raha hai
kya hota hai
Energy required to take out an electron ji
Out of a metal
oh achcha
Kinda ionisation energy hi hai tbh
But yahan single atom nahi hai
Collisions bhi honge
Baaki atoms se
So better term
Work Function
:shocked:
So you just equate that their ratio to the ratio of E
I guess you should get it
'cause E itself is voltage too right?
Vo oppose karegi photoelectric current ko
Somehow
Ek second I feel like there's a gap here
I'll come back in a moment
It's fine alright
This would work out the same
Bas ek difference to note
You are not actually taking the ratio of work functions
It's the stopping potential
Bas kyunki that "e" that you divide by
Vo cancel out ho jayega
Chemistry detected, question rejected š„š
It's maths saar š
Number Theory samajh ke kar lo saar
šš
thats such a me thing to say
I guess the most excited person about this question is not even the guy who asked
š
idk chem just doesnt hit the same
First find the wavelength from the fact that E_0-0.06/n pH=h(frequency)-w_0
ye kya hai
Uske Baad do rhe same thing, this time pH being the variable.
wait
Solution na, I think.
That's not required imk
imo
'cause do cases given hain
A ratio can be taken from values from both cases
And equated to the ratio of stopping potentials
Which is the ratio of work function basically
Denominator me ek term variable aayega
Log vala
That's the pH term
Ooh right fair point....
how do you do this wth
Atomic plus electro ki mind games
yeah i will revisit this at night
if you guys solve do ping me
Vaise maine idea likh diya hai upar
You can revisit this once you're done
E_0-0.06/n pH + (initial work function)=E_0-0.06/n pH(final)+(final work function)
Ping kar denge koi ni before marking solved
E_0 cancels out so ye 0.06/n(delta pH)=delta(work function)
How'd you get this?
Ratio nahi hona chahiye kya?
From the above thingy, I'll have to see how to take ratio, 'cuz I can't see how we would do it š
I think the net E with the initial pH divided by the net E with the final pH
Equals work function initial divided by work function final
But work function ka ratio ki jagah hv-work function ja ratio hona chahiye na?
Ahmm haaaan
But ek second
Fir provided energy ka component kahan se laayein
If I'm understanding the question hum Na ko light se irradiate kar rahe hain and the potential difference battery se balance out ho rahi hai.
Hmm I guess
And then similar setup for another metal
Hum ulta karke stop karna chah rate hain
Current flow
I guess I get this now okay
I guess I get this now
The provided energy
In the form of voltage
Should be the same in both cases
lol
They must counteract each other
isnt this direct?
*The energy provided by light.
Hmm fair
I mean kinda but framing thodi implicit hai I would say
hmmmmmm i see i see
to karna kya hai exactly
@Phalawor
OH
achcha
made smol mistake hehe
proses
im messing up calculations
this sucks
noice thanks bro
What?š
For a moment I thought you said this sucks lmao
me messing up calculation
nahhh this is epik
I know it does fr
ahem ahem
someone else calculate this out ive gotten 3 diffrent answers for 3 diffrent attempts
and none of them are correct
š
calculator
I'm done with physical chemistry
Where am I even going wrong bruh
,rotate
17
ans?
its 142
why is there a 2?
LMAO
log [H]*[Cl] hoga na Nerst me
conc of H and Cl would be same
there H^2
okay got 142
1.41 is the ph
e0 is 0.22 that is correct
ephoton is 2.64?
2.3 + 0.34 (0.34 is Ecell ?)
this is for sodium
for potassium, 2.64 - 2.25
aapke me 2.30 likha hai
so lastly 0.39 = 0.22 - 0.06/2 log Q
Wouldn't that just be the work function?
That's was a major question what would be the energy of the photon
overall rxn is H2+ 2AgCl -> 2H+ + 2Cl-
so from that you get Q
(+2Ag :D)
that is 2.3 no?
work function of Na is 2.3
ex Vs = 0.34
which you get at pH = 1
same logic as you said H+ = Cl-
ie Q = H+ ^ 4
What is this ye kya
This is stopping potential
Which is the Cell potential
But how is it 0.34
By Nerst it should 0.22 -0.06
(Sorry if I'm completely wrong this is just me 17-18th question)
First principles :)
did this make sense?
oh haa right right ;-;
sorry sorry
yea couldnt figure it
yups
sorry for vanishing like that mum sent me to get pav bhaji there was practically no signal there
na bro all good
haha awesome good you got great food :D
frfr
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