Question

caluclate the uncertainity in position of electrons having velocity of 8.2×10^4 m/s.​

Answer 1

$\underline{ \boxed{ \bold{ \mathfrak{ \purple{ \huge{Answer}}}}}} \\ \\ \implies\rm \: \triangle{v} = 8.2 \times {10}^{4} \: \frac{m}{s} \\ \\ \implies \rm \: mass \: of \: electron = 9.109 \times {10}^{ - 31} \: kg \\ \\ \star \: \red{ \mathfrak{formula}} \\ \\ \dag \: \: \underline{ \boxed{ \bold{ \pink{ \mathfrak{ \triangle{x} \times m{ \triangle{v}} = \frac{h}{4\pi} }}}}} \: \: \dag \\ \\ \star \: \red{ \mathfrak{calculation}} \\ \\ \leadsto \rm \: \triangle{x} = \frac{5 \times {10}^{ - 35} }{9.109 \times {10}^{ - 31} \times 8.2 \times {10}^{4} } \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{ \triangle{x} = 6.6 \times {10}^{ - 10} \: m}}}}} \: \purple{ \star}$

Answer 2

Answer:

7.04 x 10$^{-7}$

Explanation:

heisenberg's uncertainity rule

mass of e- =9.1 x10$^{-28}$

velocity is given