Question

If the mass of an electron reduces to one fourth of its original value, then it's new $n^{th}$ orbit radius becomes

• A. two times
• B. four times
• C. sixteen times
• D. $\frac{1}{4}$ times

Answer 1

The correct option is B four times

Using Bohr's postulate,

$mvr=\frac{nh}{2\pi }$

From the above equation we can write,

$r\propto \frac{1}{m}$

$\Rightarrow \frac{r_1}{r_2}=\frac{m_2}{m_1}$

Given, $m_2=\frac{1}{4}{m}_1$

Therefore, $\frac{r_1}{r_2}=\frac{1}{4}$

So, $r_2=4r_1$

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