An electron in the hydrogen atom jumps from the first excited state to the ground state. What will be the percentage change in the speed of the electron (in $\%$)?

The potential energy of a proton and an electron in a hydrogen atom is given by $V = V_0 \ln \left( \frac{r}{r_0} \right)$,where $r_0$ is a constant. Assuming the system follows the Bohr model,find the relationship between the radius $r_n$ and the principal quantum number $n$.

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

The difference between the $n^{th}$ and $(n+1)^{th}$ Bohr radius of an $H$ atom is equal to its $(n-1)^{th}$ Bohr radius. The value of $n$ is:

Four electrons,each of mass $m_{e}$,are in a one-dimensional box of size $L$. Assume that the electrons are non-interacting,obey the Pauli exclusion principle,and are described by standing de Broglie waves confined within the box. Define $\alpha = h^{2} / 8 m_{e} L^{2}$ and $U_{0}$ to be the ground state energy. Then,

The ratio of the radius of the first Bohr orbit to that of the second Bohr orbit of the orbital electron is