MCQ
If $m$ and $e$ are the mass and charge of the revolving electron in the orbit of radius $r$ for hydrogen atom, the total energy of the revolving electron will be
  • A
    $\frac{1}{2}\,\frac{{{e^2}}}{r}$
  • B
    $ - \frac{{{e^2}}}{r}$
  • C
    $\frac{{m{e^2}}}{r}$
  • $-\frac{1}{2}\,\frac{{{e^2}}}{r}$

Answer

Correct option: D.
$-\frac{1}{2}\,\frac{{{e^2}}}{r}$
d
Total energy of a revolving electron is the sum of its kinetic and potential energy. Total energy $= K.E. + P. E.$

$ = \frac{{{e^2}}}{{2r}} + \left( { - \frac{{{e^2}}}{r}} \right) =  - \frac{{{e^2}}}{{2r}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Which of following reaction does not involve either oxidation or reduction
The discovery that chlorofluorocarbons initiate a radical $-$ chain reaction that catalytically destroys ozone won the $1995$ Nobel prize in chemistry Given the initiation step below which of the following reactions is the most likely to serve as one of the propagation steps?
$\mathrm{CCl}_2 \mathrm{~F}_2+\mathrm{hv} \rightarrow{ }^* \mathrm{CCIF}_2+{ }^* \mathrm{Cl}$
For the reaction; $PCl_{5(g)}  \rightleftharpoons  PCl_{3(g)} +Cl_{2(g)}$ the forward reaction at constant temperature is not favoured by
The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of $LiF$ in water is due to:
Identify the incorrect statement from the following.
An electron instially present in an excited state of $H$ atom is further excited to another energy level by an incident photon. It releases $10$ photons while coming back to ground state out of which $7$ have higher energy than incident photon. The electron was instially present in
The molecule which is pyramid shape is
Which of these molecules have non $-$ bonding electron pairs on the central atom? $\ce{I : SF_4, II : ICl_3​, III : SO_2}$​
Which statement is correct about $CO_2$. If central atom use $s + P_x$ orbitals in hybridisation
$(i)$ $\pi $ bond will be formed by $P_Y$ and $P_Z$
$(ii)$ molecule can be in $XY$ plane
$(iii)$ molecule can be in $XZ$ plane
$(iv)$ molecule can be in infinite plane
$(v)$ molecule can be in $YZ$ plane
Correct code is
$H - C \equiv C\mathop {-} \limits^a C \equiv C\mathop {-} \limits^b C{H_3}$

Compare the bond lengths $a$ and $b$