Question

The energies of macroscopic objects, as well as those of microscopic objects, are quantized, but the effects of the quantization are not seen because the difference in energy between adjacent states is so small. Apply Bohr’s quantization of angular momentum to the revolution of Earth (mass$6.0\times {10}^{24}\mathrm{kg}$ ), which moves with a speed of$3.0\times {10}^4{\mathrm{ms}}^{-1}$a circular orbit (radiusrole="math" localid="1663411980771" $1.5\times {10}^{11}\mathrm{m}$ ) about the sun. The sun can be treated as fixed. Calculate the value of the quantum number n for the present state of the Earth-sun system. What would be the effect of an increase in n by 1?

Short answer

The value of n is$2.56\times {10}^{74}$ .

When the value of n changes by 1 then it would result in a change in the distance between the earth and the sun by ${10}^{74}$.

Step-by-step solution

Expression of angular momentum

The expression to calculate angular momentum is as follows:

$\mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}\to \mathrm{n}=\frac{2\mathrm{\pi mvr}}{\mathrm{h}}$

Where,

n = quantum number

m = mass of the electron

v = velocity of the electron

h = Planck’s constant

r = radius of the nth orbit

Given data

• The value of n is $6.626\times {10}^{-34}\text{J}·\text{s}$.
• The value of m is$6.0\times {10}^{24}\text{kg}$ .
• The value of v is $3.0\times {10}^4\text{m}·{\text{s}}^{-1}$.
• The value of h is 3.14.
• The value of r is $1.5\times {10}^{11}\text{m}$.

Substitute the values in the above equation.

The known values are substituted.

$$\begin{gathered} \mathrm{n}=\frac{2\mathrm{\pi mvr}}{\mathrm{h}} \\ =\frac{2(3.14)\left(6.0\times {10}^{24}\mathrm{kg}\right)\left(3.0\times {10}^4\mathrm{m}-{\mathrm{s}}^{-1}\right)\left(1.5\times {10}^{11}\mathrm{m}\right)}{6.626\times {10}^{-34}\mathrm{J}-\mathrm{s}} \\ =2.56\times {10}^{74} \end{gathered}$$

When the value ofnchanges by 1, then it would result in a change in the distance between the earth and the sun by${10}^{74}$ .