Question

In hydrogen atom, energy of first excited state is -3.4 eV. Then, KE of same orbit of hydrogen atom :-

• A. +3.4 eV
• B. +6.8 eV
• C. -13.6 eV
• D. +13.6 eV

Answer 1

Answer: (A)

+3.4 eV

Answer 2

In hydrogen atom, energy of first excited state is -3.4 eV

Then, KE of same orbit of hydrogen atom can be determined by:

$\because$ Total energy $(E_n)$= $KE + PE$
In first excited state$=\frac{1}{2}mv^2+\Bigg[-\frac{Ze^2}{r}\Bigg]$
\hspace15mm =+\frac{1}{2}\frac{Ze^2}{r}-\frac{Ze^2}{r}

$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, - 3.4 eV =-\frac{1}{2}\frac{Ze^2}{r}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, KE=\frac{1}{2}\frac{Ze^2}{r}$
= $+\, 3.4 \,eV$

So, Energy of first excited state is 3.4 eV

The kinetic energy of a particle is the form of energy that it possesses due to the motion of the particle. It is also defined as the work done in accelerating a particle of a given mass from rest to its stated velocity.

• The motion of electrons, waves, atoms, substances, molecules, and objects are examples of kinetic energy.
• A system of bodies that is the combination of many individual bodies may have internal kinetic energy due to the motion of individual bodies.
• The SI unit of kinetic energy is joule (J).
• The kinetic energy of a body of mass m moving with velocity v is given by
  K.E = 1/2 mv2

Discover more from this chapter:Structure of Atom