Question

The radius of first Bohr's orbit of hydrogen atom is $0.53\dot{A}$. Then the radius of first Bohr's orbit of Mesonic atom [negative meson has mass $207$ times that of electron but same charge] is :

• A. $2.56\times {10}^{-13}\text{m}$
• B. $1.06\times {10}^{-13}\text{m}$
• C. $0.53\times {10}^{-10}\text{m}$
• D. $7.0\times {10}^{-12}\text{m}$

Answer 1

The correct option is A $2.56\times {10}^{-13}\text{m}$

From Bohr's model, we know that the radius of hydrogen like atom is

$r=\frac{n^2{h}^2{\epsilon }_0}{\pi Z{e}^2{m}_e}$

$\Rightarrow r\propto \frac{n^2}{m_e}$

For first orbit, $n=1$

$\Rightarrow \frac{r}{r_m}=\frac{m_m}{m_e}$

Given:
$m_m=207m_e;r=0.53\dot{A}$

$\Rightarrow \frac{0.53\dot{A}}{r_m}=\frac{207m_e}{m_e}$

$\therefore r_m=0.00256\dot{A}=2.56\times {10}^{-13}\text{m}$

Hence, option $\left(\text{A}\right)$ is correct.