For an electron- if the uncertainty in velocity is - V- the uncertainty in its position - X is given by-

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Heisenberg's Uncertainty Principle

For an electr...

Question

For an electron, if the uncertainty in velocity is $Δ V$ , the uncertainty in its position $(Δ X)$ is given by:

\frac{h m}{4 X Δ V}

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\frac{2 π}{h m Δ V}

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\frac{h}{4 π m Δ V}

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\frac{2 π m}{h Δ V}

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Solution

The correct option is C $\frac{h}{4 π m Δ V}$
According to the Heisenberg's uncertainty principle, it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.
$△ x . △ p_{x} \geq \frac{h}{4 π}$

$△ x . m △ v \geq \frac{h}{4 π}$

$△ x \geq \frac{h}{4 π m △ v}$

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For an electron, if the uncertainty in velocity is ΔV, the uncertainty in its position (ΔX) is given by:

The correct option is C h4πmΔVAccording to the Heisenberg's uncertainty principle, it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.△x.△px≥h4π△x.m△v≥h4π△x≥h4πm△v

For an electron, if the uncertainty in velocity is $Δ V$ , the uncertainty in its position $(Δ X)$ is given by: