The average time between successive collisions, \( t_{\text{avg}} \), for an atom in an ideal gas depends on the mean speed of the atoms, which is proportional to the square root of the temperature \( T \).
The mean speed, \( v_{\text{mean}} \), is given by:
\[ v_{\text{mean}} \propto \sqrt{T} \]
The average time between collisions, \( t_{\text{avg}} \), is inversely proportional to the collision frequency. Since the collision frequency depends on the mean speed of the atoms, we have:
\[ t_{\text{avg}} \propto \frac{1}{v_{\text{mean}}} \]
Substituting \( v_{\text{mean}} \propto \sqrt{T} \):
\[ t_{\text{avg}} \propto \sqrt{T} \]
Thus, \( t_{\text{avg}} \) is directly proportional to \( \sqrt{T} \).
The motion of a particle in the XY plane is given by \( x(t) = 25 + 6t^2 \, \text{m} \); \( y(t) = -50 - 20t + 8t^2 \, \text{m} \). The magnitude of the initial velocity of the particle, \( v_0 \), is given by:
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)