Question

what could be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 Volts whose uncertainty in position is (150\π) nm
Given that lamda = √ 150/volt angstrome

A) 0.22
B) 6.2
C) 0.42
D) 8.2



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Answer 1

Answer:

Explanation:de broglie wavelength is,

λ=hp

from here we get

Δλ=hΔpp2 (- ve sign can be ignored here)

we know ΔxΔp≃h

so, Δp≃hΔx

and the electron is accelerated by V volt, hence the relation eV=p22mgiving p2=2meV where m is the electron mass and e is electron charge. gathering it all we land onto,

Δλ=h22meVΔx

i guess this is it. i don’t have any calculator nearby so not giving the numerical result here. values of all the variables are given already.

Thanks and please mark it as the brainliest.