Question

In accordance with the Bohr's model the quantum number that characterises the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} m$ with orbital speed $3 \times 10^{4} m / s$ (Mass of earth $=6.0 \times 10^{24} kg$ ). They belong to

• A. Balmer series
• B. Lyman series
• C. Humbnery series
• D. None of these

Answer 1

Answer: (B)

Given, radius of orbit $r=1.5 \times 10^{11} m$
Orbital speed $v =3 \times 10^{4} m / s$
Mass of earth $M =6 \times 10^{24} kg$
Angular momentum, $m v r=\frac{n h}{2 \pi}$
or $n=\frac{2 \pi v r m}{h}$ [where, $n$ is the quantum number of the orbit]
$=\frac{2 \times 3.14 \times 3 \times 10^{4} \times 1.5 \times 10^{11} \times 6 \times 10^{24}}{6.63 \times 10^{-34}}$
$=2.57 \times 10^{74}$
or $n=2.6 \times 10^{74}$
Thus, the quantum number is $2.6 \times 10^{74}$ which is too large.
The electron would jump from $n=1$ to $n=3$.
$E_{3}=\frac{-13.6}{3^{2}}=-1.5\, eV$
So, they belong to Lyman series.