For an electron- the uncertainty in velocity is Delta V- the uncertainty in its position -Delta X- is given by-dfrac-2pi -hmDelta V-dfrac-hm-4XDelta V-dfrac-h -4pi mDelta V-dfrac-2pi m-hDelta V-

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Toppr Admin · Accepted Answer

According to the Heisenberg-apos-s uncertainty principle- it is impossible-to determine simultaneously- the exact-position and exact momentum -or velocity-of an electron-x25B3-x-x25B3-px-x2265-h4-x3C0-x25B3-x-m-x25B3-v-x2265-h4-x3C0-x25B3-x-x2265-h4-x3C0-m-x25B3-v

For an electron, if the uncertainty in velocity is $Δ V$ , the uncertainty in its position $(Δ X)$ is given by:
$\frac{2 π}{h m Δ V}$
$\frac{h m}{4 X Δ V}$
$\frac{h}{4 π m Δ V}$
$\frac{2 π m}{h Δ V}$

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

$△ x . m △ v \geq \frac{h}{4 π}$

$△ x \geq \frac{h}{4 π m △ v}$

For an electron, if the uncertainty in velocity is ΔV, the uncertainty in its position (ΔX) is given by:2πhmΔVhm4XΔVh4πmΔV2πmhΔV

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.△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:
$\frac{2 π}{h m Δ V}$
$\frac{h m}{4 X Δ V}$
$\frac{h}{4 π m Δ V}$
$\frac{2 π m}{h Δ V}$

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

$△ x . m △ v \geq \frac{h}{4 π}$

$△ x \geq \frac{h}{4 π m △ v}$