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

Answer:

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

Answer 2

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

Here we go ~

According to Heisenberg's uncertainty principle :

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

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

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

$\qquad \sf \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 \times 10 {}^{ - 34} }{4 \pi \cdot 10 {}^{ - 27} \times 10 {}^{ - 3} \times 10 {}^{ - 11} }$

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

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

$\qquad \sf \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{4 \times 3.14 } \times \cfrac{10 {}^{ - 34} } {10 {}^{ - 41} }$

$\qquad \sf \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{4 \times 3.14 } \times {10 {}^{ 7}}$

$\qquad \sf \dashrightarrow \: \Delta v \geqslant \cfrac{6.6 }{12.56 } \times {10 {}^{ 7}}$

$\qquad \sf \dashrightarrow \: \Delta v \geqslant0.525\times {10 {}^{ 7}}$

$\qquad \sf \dashrightarrow \: \Delta v \geqslant5.25\times {10 {}^{ 6}} \:\:m/s$