The angular momentum of electron in 3rd Bohr orbit of Hydrogen atom is `3.165 xx 10^(-34)kg m^2` /s. Calculate Plank’s constant h.

Answer

To calculate Planck's constant \( h \) using the angular momentum of the electron in the 3rd Bohr orbit of the hydrogen atom, we can use the formula for angular momentum in Bohr's model:

\[ L = n \frac{h}{2\pi} \]

where:
- \( L \) is the angular momentum,
- \( n \) is the principal quantum number (which is 3 for the 3rd orbit),
- \( h \) is Planck's constant.

Given that the angular momentum \( L \) is \( 3.165 \times 10^{-34} \, \text{kg m}^2/\text{s} \), we can rearrange the formula to solve for \( h \):

\[ h = L \cdot \frac{2\pi}{n} \]

Now, substituting the values into the equation:

1. Substitute \( L = 3.165 \times 10^{-34} \, \text{kg m}^2/\text{s} \) and \( n = 3 \):

\[ h = 3.165 \times 10^{-34} \cdot \frac{2\pi}{3} \]

2. Calculate \( \frac{2\pi}{3} \):

\[ \frac{2\pi}{3} \approx 2.0944 \]

3. Now substitute this value back into the equation for \( h \):

\[ h = 3.165 \times 10^{-34} \cdot 2.0944 \]

4. Perform the multiplication:

\[ h \approx 6.634 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

Thus, the value of Planck's constant \( h \) is approximately:

\[ h \approx 6.634 \times 10^{-34} \, \text{J s} \]