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 \( \sqrt{r}. \)
An electron in the hydrogen atom initially in the fourth excited state makes a transition to \( n^{th} \) energy state by emitting a photon of energy 2.86 eV. The integer value of n will be 1cm.
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: