Question

Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7*10^5

Answer 1

Answer:

$\displaystyle{\Delta x\geq1.01\times10^{-10}}$

Explanation:

$\displaystyle{Given \ ;}\\\\\\\displaystyle{Velocity(v)=5.7\times10^5 \ m/sec}\\\\\\\displaystyle{We \ have \ small \ of \ electron=9.1\times10^{-31} \ kg}\\\\\\\displaystyle{Also \ value \ of \ planck's \ constant \ h=6.626\times10^{-34} \ J \ sec}\\\\\\\displaystyle{we \ know \ uncertainty \ equation}\\\\\\\displaystyle{\Delta x . \ m\Delta v\geq\frac{h}{4\pi}}\\\\\\\displaystyle{putting \ value \ in \ equation \ we \ get}$

$\displaystyle{5.7\times10^5\times9.1\times10^{-31}\times\Delta x\geq\frac{6.626\times10^{-34}}{4\pi}}\\\\\\\displaystyle{\Delta x\geq\frac{6.626\times10^{-34}}{4\times3.14\times5.7\times10^5\times9.1\times10^{-31}}}\\\\\\\displaystyle{\Delta x\geq\dfrac{6.626\times10^{-8}}{4\times3.14\times5.7\times9.1}}\\\\\\\displaystyle{\Delta x\geq\dfrac{6.626\times10^{-8}}{651.48}}\\\\\\\displaystyle{\Delta x\geq0.0101\times10^{-8}}\\\\\\\displaystyle{\Delta x\geq1.01\times10^{-10}}$

Thus we get uncertainty in the position of an electron .