Question

=> The approximate mss of an electron is 1/10²⁷ g , Calculate the uncertainty in its velocity if the uncertainty in its position were of the order of 1/10¹¹ m ( h = 6.6 × 1/10³⁴ kg m² per sec ) ..​

Answer 1

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Here we go ~

According to Heisenberg's uncertainty principle :

`\qquad \sf  \dashrightarrow \: \Delta x \Delta p \geqslant \cfrac{h}{4 \pi}`

`\qquad \sf  \dashrightarrow \: \Delta x \cdot m \Delta v \geqslant \cfrac{h}{4 \pi}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{h}{4 \pi \cdot m \Delta x}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 * 10 {}^( - 34) }{4 \pi \cdot 10 {}^( - 27) * 10 {}^( - 3) * 10 {}^( - 11) }`

`\textsf{ [ we took }`
`{ {10}^(-3) }`
`\textsf{because mass of electron is}`
`\textsf{ given in grams that need to be converted into kg ]}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{4 \pi } * \cfrac{10 {}^( - 34) } {10 {}^( (- 27 - 3 - 11)) }`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{4 * 3.14 } * \cfrac{10 {}^( - 34) } {10 {}^( - 41) }`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{4 * 3.14 } * {10 {}^( 7)}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{12.56 } * {10 {}^( 7)}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant0.525* {10 {}^( 7)}`

`\qquad \sf  \dashrightarrow \: \Delta v \geqslant5.25* {10 {}^( 6)} \:\:m/s`

Answer 2

Δx(m.Δv)=h/4π

here ,

Δv = uncertainty in velocity

10−11×10−27×Δv=6.626×10−34/4×22/7

=5.25×103ms−1