According to Bohr’s model of the hydrogen atom, the angular momentum \( L \) of an electron in an orbit is quantized and given by:
\( L = n\hbar, \)
where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck’s constant.
For a hydrogen atom, the radius of the \( n \)-th orbit is given by:
\( r_n \propto n^2. \)
Therefore, we can express \( n \) in terms of \( r \):
\( n \propto \sqrt{r}. \)
Substituting this into the expression for angular momentum:
\( L \propto n \propto \sqrt{r}. \)
Hence, the angular momentum of an electron in a hydrogen atom is proportional to \( \sqrt{r}. \)
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).