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 × 1011 m with orbital speed 3 × 104 ms -1 is: (Take mass of the earth .=6 × 1024 kg )

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 × 1011 m with orbital speed 3 × 104 ms -1 is: (Take mass of the earth .=6 × 1024 kg ) (A) 5.98 × 1086 (B) 2.57 × 1038 (C) 8.57 × 1064 (D) 2.57 × 1074

Tardigrade Admin · Accepted Answer

According to Bohr's model mvr =( nh /2 π) n =(2 π mvr / h )=(2 × 22 × 6 × 1024 × 3 × 104 × 1.5 × 1011/7 × 6.6 × 10-34) n =(2 × 6 × 15 × 1073/7) n =2.57 × 1074

Q. In accordance with the Bohr's model, the quantum number that characterises the earth's revolution around the sun in an orbit of radius $1.5 \times 1 0^{11} m$ with orbital speed $3 \times 1 0^{4} m s^{- 1}$ is: (Take mass of the earth $= 6 \times 1 0^{24} k g)$

$5.98 \times 1 0^{86}$

$2.57 \times 1 0^{38}$

$8.57 \times 1 0^{64}$

$2.57 \times 1 0^{74}$

According to Bohr's model $m v r = \frac{nh}{2 π}$
$n = \frac{2 πm v r}{h} = \frac{2 \times 22 \times 6 \times 1 0 ^{24} \times 3 \times 1 0 ^{4} \times 1.5 \times 1 0 ^{11}}{7 \times 6.6 \times 1 0 ^{- 34}}$
$n = \frac{2 \times 6 \times 15 \times 1 0 ^{73}}{7}$
$n = 2.57 \times 1 0^{74}$

Q. In accordance with the Bohr's model, the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5×1011m with orbital speed 3×104ms−1 is: (Take mass of the earth =6×1024kg)

5.98×1086

2.57×1038

8.57×1064

2.57×1074