Atomic Structure: De - Broglie Wavelength
Doubt in question 7.
My approach: I used the formula λ = h/2mqV.
My answer: V = 4 * 10^34 Volts (☠️)
Actual answer: V = 32.86 Volts
36 Replies
@Dexter
Note for OP
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p= root (2m E) E=kinetic energy
h/lamda= root (2m E) and E= eV (V= appliead potential differnce)
h/lamda= root (2meV)
isse lamda ki sahi value aa rhi?
Lamda toh given hai na
accha toh V ki sahi value aa rhi kya
V ki value 4 * 10³⁴ Volts 💀
matlab sahi nahi a rha is expression se calculate karne par?
Naw, you can check my calculation though
then idk XD
acdelerating potential ka matlab potential difference hi hau na
Yeah
Name checks out XD
Just kidding
arey ye toh modern physics mein bhi hai 🤡
Ya in 12th
,rotate
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Here
whats the answer?
32.86 Volts
lamda= h/mv
find v
then eV= 1/2 mv^2
Yeah thats how all the websites have answered this question
What's wrong with my approach?
send the whole working
Sir hamare teacher bilkul theory nahi padhate, they just make the students mug up all the formulas
For me I was studying side by side from a teacher online and he taught the theory well so my theory part is good.
In this question, my only doubt is why I should find velocity first and then the voltage when I have this formula with all the values compiled in itself.
Does acceleration potential mean something related to velocity and voltage both?
Agar is question me sirf potential pucha hota to kya mai λ = h/2mqV use kar sakta tha?
dude
if you solve by this method you'll get 32.27 which is pretty close
i directly wrote some stuff cuz calc was lengthy asf
@Sidd
Oh shoot
I forgot to square plancks constant
All this ruckus due to a calculation mistake
I am sorry guys
its okay 🙂
I'll try again
Got the answer
Thanks a lot guys
+solved @trin
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