To calculate the moment of inertia \(I\) for a diatomic molecule such as hydrogen (\(H_2\)), use the formula:
\[I = \mu r^2\]
where \(\mu\) is the reduced mass given by \(\mu = \frac{m_1 \times m_2}{m_1 + m_2}\) for two identical masses \(m\), and \(r\) is the inter-molecular distance. For two protons, each with mass \(m = 1.67 \times 10^{-27} \, \text{kg}\):
\[\mu = \frac{m \times m}{2m} = \frac{m}{2} = \frac{1.67 \times 10^{-27} \, \text{kg}}{2} = 0.835 \times 10^{-27} \, \text{kg}\]
\[r = 0.77 \times 10^{-10} \, \text{m} = 0.77 \, \text{\AA}\]
\[I = 0.835 \times 10^{-27} \, \text{kg} \times (0.77 \times 10^{-10} \, \text{m})^2 \approx 0.495 \times 10^{-47} \, \text{kg-m}^2\]
List I | List II |
---|---|
(A) (∂S/∂P)T | (I) (∂P/∂T)V |
(B) (∂T/∂V)S | (II) (∂V/∂S)P |
(C) (∂T/∂P)S | (III) -(∂V/∂T)P |
(D) (∂S/∂V)T | (IV) -(∂P/∂S)V |
Ultraviolet light of wavelength 350 nm and intensity \(1.00Wm^{−2 }\) falls on a potassium surface. The maximum kinetic energy of the photoelectron is