Question

The circumference of first Bohr orbit is $3.32\times {10}^{-10}m$. If this circumference represents one wavelength, what is the velocity of electron in this orbit?

• A. $2.2\times {10}^{6}m$ $se{c}^{-1}$
• B. $22\times {10}^{7}m$ $se{c}^{-1}$
• C. $123\times {10}^{6}m$ $se{c}^{-1}$
• D. $90\times {10}^{6}m$ $se{c}^{-1}$

Answer 1

The correct option is A $2.2\times {10}^{6}m$ $se{c}^{-1}$

The de Broglie equation is $\lambda =\frac{h}{mv}$
Hence, $v=\frac{h}{m\lambda }$
Substittue values in the above expression.
$v=\frac{6.62\times {10}^{-34}}{9.11\times {10}^{-31}\times 3.32\times {10}^{-10}}=2.2\times {10}^{6}m/s$
Hence, the velocity of the electron in the orbit is $2.2\times {10}^{6}m/s$