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 $1.5 \times 10^{11}\text{ m}$ with orbital speed$3 \times 10^{4}\text{ m }\text{s}^{- 1}\text{ is}(\text{Take mass of earth } = \text{ 6} \times 10^{24}\text{ kg})$

• A. $5.98 \times 10^{86}$
• B. $2.57 \times 10^{38}$
• C. $8.57 \times 10^{64}$
• D. $2.57 \times 10^{74}$

Answer 1

Answer: (D)

$2.57 \times 10^{74}$

Answer 2

Here $r = 1.5 \times 10^{11}m$

$v = 3 \times 10^{4}ms^{- 1}$

$m = 6.0 \times 10^{24}kg$

According to Bohr’s model

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

$\Rightarrow n = \frac{2\pi mvr}{h}$

$= \frac{2 \times 22}{7} \times \frac{6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{6.6 \times 10^{- 34}}$

$= 2.57 \times 10^{74}$