In an atom,an electron is moving with a speed | vedclass

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(B) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4 \pi}$.Since $\Delta p = m 	imes \Delta v$,the formula b

Vedclass · Accepted Answer

$1.92 	imes 10^{-3} \ m$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4 \pi}$.Since $\Delta p = m 	imes \Delta v$,the formula becomes $\Delta x = \frac{h}{4 \pi 	imes m 	imes \Delta v}$.Given: $v = 600 \ m/sec$,uncertainty in velocity $\Delta v = 600 	imes \frac{0.005}{100} = 0.03 \ m/sec$.Mass of electron $m = 9.10 	imes 10^{-31} \ kg$,Planck's constant $h = 6.626 	imes 10^{-34} \ J \cdot s$.Substituting the values:$\Delta x = \frac{6.626 	imes 10^{-34}}{4 	imes 3.1416 	imes 9.10 	imes 10^{-31} 	imes 0.03}$.$\Delta x = \frac{6.626 	imes 10^{-34}}{3.424 	imes 10^{-31}} \approx 1.93 	imes 10^{-3} \ m$.Rounding to the nearest provided option,the correct answer is $1.92 	imes 10^{-3} \ m$.

In an atom,an electron is moving with a speed of $600 \ m/sec$ with an uncertainty of $0.005\%$. The uncertainty with which the position of the electron can be located is:

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