Question

What will be the minimum uncertainty in velocity of a particle of mass $1.1\times {10}^{-28}$ kg if uncertainty in its position is $3\times {10}^{-10}$ cm? $(h=6.62\times {10}^{-34}kg{m}^2{s}^{-1})$

• A. $1.6\times {10}^{5}m{s}^{-1}$
• B. $2.6\times {10}^{5}m{s}^{-1}$
• C. $3.6\times {10}^{5}m{s}^{-1}$
• D. $6.1\times {10}^{5}m{s}^{-1}$

Answer 1

The correct option is A $1.6\times {10}^{5}m{s}^{-1}$

According uncertainity principle,

$\Delta x.\Delta p=h/4\pi$

$\Delta x(m.\Delta v)=h/4\pi$

$\Delta v=\left(h/4\pi \right)\times (1/m.\Delta x)$

$=\frac{6.62\times {10}^{-34}}{4\times 3.14}\times \frac{1}{(1.1\times {10}^{-28})(3\times {10}^{-12})}$ $m{s}^{-1}$

$=1.6\times {10}^{5}m{s}^{-1}$

Hence, the correct option is $A$