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The correct expression derived for the energy of an electron in the nth energy level is - (in CGS)
( (A) E_{n}=frac{2 pi^{2} m e^{4}}{n^{2} h^{2}} )
( (B) E_{n}=-frac{2 pi^{2} m e^{4}}{n h^{2}} )
(C) ( E_{n}=-frac{2 pi^{2} mathrm{me}^{2}}{mathrm{n}^{2} mathrm{h}^{2}} )
( (D) E_{n}=-frac{2 pi^{2} m e^{4}}{n^{2} h^{2}} )

JEE/Engineering Exams
Physics
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108
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The correct expression derived for the energy of an electron in the nth energy level is - (in CGS)
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When electron is revolving around the electron, it has kinetic energy. since. it is attracted towards the nucleus by electrostatic force of attraction, it has potential energy. Therefore, the sum of kinetic energy and potential energy gives the total energy.
We know, Kinetic energy of electron in nth orbit is given by
[
K E=frac{1}{2} m v_{n}^{2} ldots ldots . .(i)
]
We know from previous topic.
[
v_{n}=frac{e^{2}}{2 varepsilon_{o} n h}
]
So, putting the value of ( mathrm{V}_{mathrm{n}} ) in equation (i), we get
[
begin{array}{l}
K E=frac{1}{2} mleft(frac{e^{2}}{2 varepsilon_{0} n h}right)^{2}
therefore K E=frac{m e^{4}}{8 varepsilon_{0}^{2} n^{2} h^{2}}
end{array}
]
Similarly, Potential energy (P.E.) = Potential due to nucleus at distance r x charge of electron i.e.
[
begin{array}{l}
P E=frac{e}{4 pi varepsilon_{o} r_{n}} times(-e)
text { or, } P E=-frac{1}{4 pi varepsilon_{circ}} frac{e^{2}}{r_{n}}
end{array}
]
We know from the previous topic.
[
r_{n}=frac{varepsilon_{o} n^{2} h^{2}}{pi m e^{2}}
]
Then Potential energy becomes,
[
begin{array}{c}
P E=-frac{e^{2}}{4 varepsilon_{0}} frac{pi m e^{2}}{varepsilon_{0} n^{2} h^{2}}
therefore P E=-frac{m e^{4}}{4 varepsilon_{0}^{2} n^{2} h^{2}}
qquad begin{array}{c}
o r, E_{n}=frac{m e^{4}}{8 varepsilon_{0}^{2} n^{2} h^{2}}-frac{m e^{4}}{4 varepsilon_{0}^{2} n^{2} h^{2}}
text { Now, Total Energy }left(E_{n}right)=text { Kinetic Energy }+text { Potential Energy }
therefore E_{n}=frac{m e^{4}}{8 varepsilon_{0}^{2} n^{2} h^{2}}
end{array}
end{array}
]

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