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The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18}$ g cm s$^{-1}$. The uncertainty in electron velocity is :
(Mass of an electron is $9 \times 10^{-28}$ g)
(Mass of an electron is $9 \times 10^{-28}$ g)
A
$1 \times 10^{5}$ cm s$^{-1}$
B
$1 \times 10^{11}$ cm s$^{-1}$
C
$1 \times 10^{9}$ cm s$^{-1}$
D
$1 \times 10^{6}$ cm s$^{-1}$
Detailed Solution
Momentum $p = mv$, so uncertainty in momentum $\Delta p = m\,\Delta v$
Uncertainty in momentum $(m\,\Delta v) = 1 \times 10^{-18}$ g cm s$^{-1}$
Uncertainty in velocity: $\Delta v = \dfrac{\Delta p}{m}$
$\Delta v = \dfrac{1 \times 10^{-18}}{9 \times 10^{-28}}$
$\Delta v = 1.1 \times 10^{9}$ cm s$^{-1}$
$\approx 1 \times 10^{9}$ cm s$^{-1}$
Uncertainty in momentum $(m\,\Delta v) = 1 \times 10^{-18}$ g cm s$^{-1}$
Uncertainty in velocity: $\Delta v = \dfrac{\Delta p}{m}$
$\Delta v = \dfrac{1 \times 10^{-18}}{9 \times 10^{-28}}$
$\Delta v = 1.1 \times 10^{9}$ cm s$^{-1}$
$\approx 1 \times 10^{9}$ cm s$^{-1}$