The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is proportional to ($e=$ electronic charge).

  • A
    $\frac{e^2}{4 r}$
  • B
    $\frac{e^2}{2 r}$
  • C
    $\frac{e^2}{r}$
  • D
    $\frac{e^2}{2 r^2}$

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Similar Questions

In a mixture of $H-He^{+}$ gas ($He^{+}$ is a singly ionized $He$ atom),$H$ atoms and $He^{+}$ ions are excited to their respective first excited states. Subsequently,$H$ atoms transfer their total excitation energy to $He^{+}$ ions by collisions. Assume that the Bohr model of the atom is exactly valid.
$1.$ The quantum number $n$ of the state finally populated in $He^{+}$ ions is
$(A) 2$ $(B) 3$ $(C) 4$ $(D) 5$
$2.$ The wavelength of light emitted in the visible region by $He^{+}$ ions after collisions with $H$ atoms is
$(A) 6.5 \times 10^{-7} \ m$ $(B) 5.6 \times 10^{-7} \ m$ $(C) 4.8 \times 10^{-7} \ m$ $(D) 4.0 \times 10^{-7} \ m$
$3.$ The ratio of the kinetic energy of the $n=2$ electron for the $H$ atom to that of the $He^{+}$ ion is
$(A) 1/4$ $(B) 1/2$ $(C) 1$ $(D) 2$

The frequency of revolution of an electron revolving in the $n^{th}$ orbit of an $H$-atom is proportional to:

$A$ moving Hydrogen atom makes a head-on collision with a stationary Hydrogen atom. Before the collision,both atoms are in the ground state,and after the collision,they move together. The minimum kinetic energy of the moving Hydrogen atom,such that one of the atoms reaches the excitation state,is (in $eV$)

In a hydrogen-like ion,the energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \ eV$. The atomic number of the ion is:

If $E$ and $L$ denote the magnitude of total energy and angular momentum of a revolving electron in the $n^{th}$ Bohr orbit, then:

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