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 $3 \times 10^{4} ms ^{-1}$ is: (Take mass of the earth $\left.=6 \times 10^{24} kg \right)$

• 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)

According to Bohr's model $mvr =\frac{ nh }{2 \pi}$
$n =\frac{2 \pi mvr }{ h }=\frac{2 \times 22 \times 6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{7 \times 6.6 \times 10^{-34}}$
$n =\frac{2 \times 6 \times 15 \times 10^{73}}{7}$
$n =2.57 \times 10^{74}$