Question

The speed of electrons in the second orbit of $B{e}^{3+}$ ion will be

• A. $\frac{c}{137}$
• B. $\frac{2c}{137}$
• C. $\frac{3c}{137}$
• D. $\frac{4c}{137}$

Answer 1

Answer: (B)

$\frac{2c}{137}$

Here, $Z=4,n=2$
Therefore, speed is $=2.18\times {10}^6\times \frac{4}{2}=4.36\times {10}^{6}m/s$
Value of $c$ is $3\times {10}^{8}m/s$.
Therefore, answer is $\frac{2c}{137}$.