Question

The uncertainty for the calculation of the radius of the $1^{st}$ Bohr orbit is 2% for the hydrogen atom. What will be the minimum uncertainty in velocity of an electron in the $1^{st}$ Bohr orbit?

$(h=6.626\times {10}^{–34}J.s,m_e=9.1\times {10}^{–31}kg)$

• A. $0.47\times {10}^{7}m/s$
• B. $1.47\times {10}^{2}m/s$
• C. $5.47\times {10}^{7}m/s$
• D. $5.47\times {10}^{2}m/s$

Answer 1

The correct option is C $5.47\times {10}^{7}m/s$

$r=0.529\times {10}^{-10}m.$
If the uncertainty in radius is 2%,
$△x=0.529\times {10}^{-10}\times \frac{2}{100}m=1.06\times {10}^{-12}m.$

Now, $△x.m.△v\ge \frac{h}{4\pi }$

$△v\ge \frac{6.626\times {10}^{-34}}{4\times 3.14\times 9.1\times {10}^{-31}\times 1.06\times {10}^{-12}}$

$△v\ge 5.47\times {10}^{7}m/s$