For an electron, if the uncertainty in velocity is Δ v , the uncertainty in position Δ x is given by:

Question

For an electron, if the uncertainty in velocity is Δ v , the uncertainty in position Δ x is given by: (A) ( hm /4 π Δ v ) (B) (4 π/ hm Δ v ) (C) ( h /4 π m Δ v ) (D) (4 π m / h Δ v )

Tardigrade Admin · Accepted Answer

According to Heisenberg's uncertainty principle Δ x ⋅ Δ p ≥ (h/4 π) But Δ p (momentum) = m Δ v Δ x ⋅ m Δ v ≥ ( h /4 π) Δ x =( h /4 π m Δ v )

Q. For an electron, if the uncertainty in velocity is $Δ v,$ the uncertainty in position $Δ x$ is given by:

$\frac{hm}{4 π Δ v}$

$\frac{4 π}{hm Δ v}$

$\frac{h}{4 πm Δ v}$

$\frac{4 πm}{h Δ v}$

According to Heisenberg's uncertainty principle

$Δ x \cdot Δ p \geq \frac{h}{4 π}$

But $Δ p$ (momentum) $= m Δ v$

$Δ x \cdot m Δ v \geq \frac{h}{4 π}$

$Δ x = \frac{h}{4 πm Δ v}$

Q. For an electron, if the uncertainty in velocity is Δv, the uncertainty in position Δx is given by:

4πΔvhm​

hmΔv4π​

4πmΔvh​

hΔv4πm​