JEE MAIN - Chemistry (2021 - 25th July Evening Shift - No. 17)
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]
[Use mass of electron = 9.1 $$\times$$ 10$$-$$31 kg, h =6.63 $$\times$$ 10$$-$$34 Js, $$\pi$$ = 3.14]
Answer
58
Explanation
$$\Delta v = {{0.02} \over {100}} \times 5 \times {10^6} = {10^3}$$ m/s
$$\Delta x.\Delta v = {h \over {4\pi m}}$$
$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$$
$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$$
$$ \Rightarrow $$ $$x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$$
$$\Delta x.\Delta v = {h \over {4\pi m}}$$
$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$$
$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$$
$$ \Rightarrow $$ $$x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$$
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