Question

Find the value of principal quantum number n corresponding 20 kg satellite.

• A. $1.6\times {10}^{45}$
• B. $7.6\times {10}^5$
• C. $1.6\times {10}^8$
• D. $8.3\times {10}^{11}$

Answer 1

Answer: (A)

We know that:

$nh=mvR$, where $R=$ $8060\times {10}^{3}m$

Also, $v=\sqrt{\frac{GM}{R}}$, $M=$ mass of earth $=6\times {10}^{2}4kg$

$n=\frac{m}{h}\sqrt{\frac{GM}{R}}R=\frac{20}{6.62\times {10}^{-34}}\sqrt{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 6400\times {10}^3}8060\times {10}^3$

$=\frac{20\times {10}^{34}}{6.62}\times {10}^9\times 8\times \sqrt{40}=1.6\times {10}^{45}$