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The question says: A proton is accelerated to one tenth of the velocity of light. If it's velocity can be measured with a precision $\pm 1\%$. What must be its uncertainty in position?

Therefore,

$v=0.1\cdot c =3\cdot 10^7\:\mathrm{m/s}\\ \Delta(v)=\frac{1}{100}\\ m= 1.6\cdot10^{-27}\:\mathrm{kg}$

Then I directly substituted these values into the formula: $\Delta(v)\cdot\Delta(\mathrm{position})\cdot \mathrm{mass}=\frac h{4\pi}$

To get the uncertainty in position, however, the answer I got was approx $3.5 \cdot 10^{-6}\:\mathrm{m}$ which is way too different from the correct answer: $0.5\cdot10^{-13}\:\mathrm{m}.$ Can anyone please explain me how to solve this question?

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1 Answer 1

up vote 4 down vote accepted

The question tells you \begin{aligned} &&v&=3\cdot10^7\:\mathrm{m/s}\pm1\%\\ &&\Delta v &= |(v+1\%v) - (v-1\%v)|\\ \implies&&\Delta v &= \frac{2}{100}v\\ \therefore&&\Delta v&= 2\cdot3\cdot10^5\:\mathrm{m/s} \end{aligned}

Hence \begin{aligned} &&\Delta x\cdot\Delta p &\geq \frac{\hbar}{2}\\ \implies&&\Delta x\cdot\Delta (v\cdot m) &\geq \frac{\hbar}{2}\\ \implies&&\Delta x&\geq \frac{\hbar}{2\cdot\Delta v\cdot m }\\ \therefore&&\Delta x &\gtrapprox\frac{1.05\cdot10^{−34}\:\mathrm{J\cdot s}}{2\cdot6\cdot10^5\:\mathrm{m/s}\cdot1.6\cdot10^{−27}\:\mathrm{kg}}\\ &&\Delta x&\gtrapprox5.5\cdot10^{-14}\:\mathrm{m}\\ \end{aligned}

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