Example 4 If the speed of electron in the first orbit of a hydrogen atom is 2cdot 10^6m/s then according to Bohr's postulate , find the radius of this orbit (pm)

To find the radius of the first orbit of a hydrogen atom according to Bohr's postulate, we can use the formula:<br /><br />\[ r = \frac{n^2 \cdot h}{4 \cdot \pi \cdot m \cdot v} \]<br /><br />where:<br />- \( r \) is the radius of the orbit,<br />- \( n \) is the principal quantum number (for the first orbit, \( n = 1 \)),<br />- \( h \) is the Planck constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),<br />- \( m \) is the mass of the electron (\( 9.109 \times 10^{-31} \, \text{kg} \)),<br />- \( v \) is the speed of the electron in the orbit (\( 2 \times 10^6 \, \text{m/s} \)).<br /><br />Substituting the given values into the formula:<br /><br />\[ r = \frac{1^2 \cdot 6.626 \times 10^{-34}}{4 \cdot \pi \cdot 9.109 \times 10^{-31} \cdot 2 \times 10^6} \]<br /><br />\[ r = \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 9.109 \times 10^{-31} \cdot 2 \times 10^6} \]<br /><br />\[ r = \frac{6.626 \times 10^{-34}}{1.146 \times 10^{-28}} \]<br /><br />\[ r = 5.78 \times 10^{-7} \, \text{m} \]<br /><br />To convert this to picometers (pm), we multiply by \( 10^{12} \):<br /><br />\[ r = 5.78 \times 10^{-7} \, \text{m} \times 10^{12} \, \text{pm/m} \]<br /><br />\[ r = 578 \, \text{pm} \]<br /><br />Therefore, the radius of the first orbit of a hydrogen atom according to Bohr's postulate is approximately 578 pm.