Question

The measurement of the electron position is
associated with an uncertainty in momentum,
which is equal to 1 x 10^-18 g cm s^-1. The
uncertainty in electron velocity is (mass of
an electron is 9 x 10^-28 g). PLEASE give a WELL EXPLAINED answer. Don't SPAM!!​

Answer 1

Answer:

$1*{10}^{6} cm {s}^{-1}$

Explanation:

As we know momentum (p) is the product of mass (m) and velocity (v)

$\Delta {P} = m \Delta {v}$

$1*{10}^{18} gcm {s}^{-1} = 9*{10}^{-28} g * \Delta {v}$

by solving we get

$\Delta {v} = 0.11* {10}^{10} cm{s}^{-1}$

therefore,

$\Delta {v} = 1.1* {10}^{9} cm{s}^{-1}$

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Answer 2

Answer:

Answer:

1*{10}^{6} cm {s}^{-1}1∗10

6

cms

−1

Explanation:

As we know momentum (p) is the product of mass (m) and velocity (v)

\Delta {P} = m \Delta {v}ΔP=mΔv

1*{10}^{18} gcm {s}^{-1} = 9*{10}^{-28} g * \Delta {v}1∗10

18

gcms

−1

=9∗10

−28

g∗Δv

by solving we get

\Delta {v} = 0.11* {10}^{10} cm{s}^{-1}Δv=0.11∗10

10

cms

−1

therefore,

\Delta {v} = 1.1* {10}^{9} cm{s}^{-1}Δv=1.1∗10

9

cms

−1