Question

The velocity of an electron of mass $9.1\times {10}^{-31}kg$ moving round the nucleus in the Bohr orbit (diameter of the orbit is $1.058\mathring{A}$ is $2.2\times {10}^{-6}m$. The magnitude of its linear momentum in this motion is

• A. $1.1\times {10}^{-34}kg-m/s$
• B. $2.0\times {10}^{-24}kg-m/s$
• C. $4.0\times {10}^{-24}kg-m/s$
• D. $4.0\times {10}^{-31}kg-m/s$

Answer 1

The correct option is B $2.0\times {10}^{-24}kg-m/s$

Momentum $p=mv=9.1\times {10}^{-31}\times 2.2\times {10}^{-6}m/s=20.02\times {10}^{-37}m/s$

$\frac{△p}{p}=\frac{1}{100}⟹△p=p\times \frac{1}{100}=20.02\times {10}^{-39}$

$\because △p\times △x=\frac{h}{4\pi }$ [Heisenberg's uncertainty principle]

$\because △x=\frac{h}{4\pi △p}=\frac{6.6\times {10}^{-34}}{4\times \pi \times 20.02\times {10}^{-39}}=2.64\times {10}^{3}m$