An accelerated electron has a speed of 5 $\times$ 106 ms$-$1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x $\times$ 10$-$9 m. The value of x is ____________. (Nearest integer)
[Use mass of electron = 9.1 $\times$ 10$-$31 kg, h =6.63 $\times$ 10$-$34 Js, $\pi$ = 3.14]
Answer (integer)
58
Solution
$\Delta v = {{0.02} \over {100}} \times 5 \times {10^6} = {10^3}$ m/s<br><br>$\Delta x.\Delta v = {h \over {4\pi m}}$<br><br>$\Rightarrow$ $$x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$$<br><br>$\Rightarrow$ $x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$<br><br>$\Rightarrow$ $x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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