Question

The velocity of an electron in a certain Bohr's orbit of $H$-atom bears the ratio $1:412$ to the velocity of light. Calculate the value of orbit number $n$ .

• A. 3
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 3

Given: $\text{Velocity of electron}=\frac{1}{412}\times \text{Velocity of light}..\left(i\right)$
We know,
${\text{v}}_n=\frac{2.189\times {10}^8\times Z}{n}\text{cm/sec}...\left(ii\right)$

Equating eq. (i) and (ii), we get

$\frac{2.189\times {10}^8\times 1}{n}=\frac{1}{275}\times (3\times {10}^{10})$
where, velocity of light c = $3\times {10}^{10}cm/s$
$\Rightarrow n=\frac{2.189\times {10}^8\times 412}{3\times {10}^{10}}=3$