Question

Speed of electron in its lst Bohr's orbit is given by $2.18 \times 10^{6} ms ^{-1}$. If the time period of electron in $n$ th orbit is measured as $4.10$ femto second, the value of $n$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

Given, speed of the electron in lst Bohr's orbit,
$v_{1}=2.18 \times 10^{6} m / s$
Time period of $n$ th orbit,
$T_{n}=4.10 f - s =4.1 \times 10^{-15} s$
Radius of Bohr's first orbit, $r_{1}=0.53 \times 10^{-10} m$
$\therefore $ Orbital period of electron in Bohr's first orbit,
$T_{1}=\frac{2 \pi r_{1}}{v_{1}}=\frac{2 \times 3.14 \times 0.53 \times 10^{-10}}{2.18 \times 10^{6}}$
$T_{1}=1.52 \times 10^{-16} s$
Again, Time period of $n$ th orbit is given by
$T_{n}=n^{3} T_{1}$
$\Rightarrow n^{3}=\frac{T_{n}}{T_{1}}=\frac{4.1 \times 10^{-15}}{1.52 \times 10^{-16}}=26.98 $
$n^{3} =27 \Rightarrow n=(27)^{1 / 3} \Rightarrow n=3$