
Using Rutherford model of atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
Answer
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Hint:A very small amount of the positively charged ions, and much of the mass of a molecule. It is known as the nucleus in this area.
The blueprint from Rutherford recommended encircling the nucleus of an atom with negative electrons. He also stated that the electrons that form the nucleus spin in circular paths at very high speed. These ring roads he named as orbits.
Electrons that are charged negatively and the nucleus is a highly packed mass of positive ions, held together by a powerful electrostatic attraction force.
Complete step by Step Solution:
According to Rutherford’s model of atom, we know that,
\[\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {{{r^2}}}\]
The mass atomic number of the atom is $Z$
The charge on the electron is $e$
The radius of the atom is $r$
The mass of the atom is $m$
The velocity is $v$
\[m{v^2} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {{{r^ {}}}}\]
Now,
We know that the total energy $T$ of the electron in an atom is defined as the sum of potential energy $P$ and kinetic energy $K$ .
$T = P + K$
$T = \dfrac {{- 1}} {{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {r} + \dfrac{1}{2}m{v^2}$
$ = \dfrac {{- 1}} {{8\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {r} $
The total energy is negative because the electron-nucleus is an attractive system.
Note:A large fraction of the $\alpha $ -particles bombarded in the direction of the golden sheet went through it, and the remainder of the area of an atom was then vacant. Any $\alpha $ -particles were bent in very narrow angles by the gold sheet, and thus the positive charge in an atom isn't spread evenly. A very small amount of the positive load is concentrated in the atom.
A few of the $\alpha $ -particles have been deflected backwards, which are only a few $\alpha $ -particles of almost ${180^0} $ deflective angles. Thus, compared with the average volume of an atom, the volume filled by the positively charged particles is very small.
The blueprint from Rutherford recommended encircling the nucleus of an atom with negative electrons. He also stated that the electrons that form the nucleus spin in circular paths at very high speed. These ring roads he named as orbits.
Electrons that are charged negatively and the nucleus is a highly packed mass of positive ions, held together by a powerful electrostatic attraction force.
Complete step by Step Solution:
According to Rutherford’s model of atom, we know that,
\[\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {{{r^2}}}\]
The mass atomic number of the atom is $Z$
The charge on the electron is $e$
The radius of the atom is $r$
The mass of the atom is $m$
The velocity is $v$
\[m{v^2} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {{{r^ {}}}}\]
Now,
We know that the total energy $T$ of the electron in an atom is defined as the sum of potential energy $P$ and kinetic energy $K$ .
$T = P + K$
$T = \dfrac {{- 1}} {{4\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {r} + \dfrac{1}{2}m{v^2}$
$ = \dfrac {{- 1}} {{8\pi {\varepsilon _0}}}\dfrac {{Z {e^2}}} {r} $
The total energy is negative because the electron-nucleus is an attractive system.
Note:A large fraction of the $\alpha $ -particles bombarded in the direction of the golden sheet went through it, and the remainder of the area of an atom was then vacant. Any $\alpha $ -particles were bent in very narrow angles by the gold sheet, and thus the positive charge in an atom isn't spread evenly. A very small amount of the positive load is concentrated in the atom.
A few of the $\alpha $ -particles have been deflected backwards, which are only a few $\alpha $ -particles of almost ${180^0} $ deflective angles. Thus, compared with the average volume of an atom, the volume filled by the positively charged particles is very small.
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