Question:
In hydrogen atom, if the kinetic energy of an electron in an orbit having angular momentum $2h/\pi$ is E, then the potential energy of the electron in the first orbit of hydrogen atom is:
In hydrogen atom, if the kinetic energy of an electron in an orbit having angular momentum $2h/\pi$ is E, then the potential energy of the electron in the first orbit of hydrogen atom is:
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For any Bohr orbit, $PE = -2 KE$.
Updated On: Jun 10, 2026
- $-32E$
- $-8E$
- $-4E$
- $-16E$
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The Correct Option is A
Solution and Explanation
Step 1: Find the orbit number.
Bohr says the angular momentum is $L=\frac{nh}{2\pi}$. Here $L=\frac{2h}{\pi}=\frac{4h}{2\pi}$, so $n=4$.
Step 2: Kinetic energy in that orbit.
For hydrogen $KE_n=\frac{13.6}{n^2}$ eV, so $KE_4=\frac{13.6}{16}$ eV, and this is the given $E$. Hence $13.6=16E$.
Step 3: Kinetic energy of the first orbit.
$KE_1=\frac{13.6}{1^2}=13.6$ eV $=16E$.
Step 4: Use the energy relation.
In any Bohr orbit $PE=-2\,KE$, because the potential energy is twice the total energy and equal to minus twice the kinetic energy.
Step 5: Potential energy of the first orbit.
$PE_1=-2\,KE_1=-2(16E)=-32E$.
Step 6: Conclusion.
The potential energy of the electron in the first orbit is $-32E$.
\[ \boxed{-32E} \]
Bohr says the angular momentum is $L=\frac{nh}{2\pi}$. Here $L=\frac{2h}{\pi}=\frac{4h}{2\pi}$, so $n=4$.
Step 2: Kinetic energy in that orbit.
For hydrogen $KE_n=\frac{13.6}{n^2}$ eV, so $KE_4=\frac{13.6}{16}$ eV, and this is the given $E$. Hence $13.6=16E$.
Step 3: Kinetic energy of the first orbit.
$KE_1=\frac{13.6}{1^2}=13.6$ eV $=16E$.
Step 4: Use the energy relation.
In any Bohr orbit $PE=-2\,KE$, because the potential energy is twice the total energy and equal to minus twice the kinetic energy.
Step 5: Potential energy of the first orbit.
$PE_1=-2\,KE_1=-2(16E)=-32E$.
Step 6: Conclusion.
The potential energy of the electron in the first orbit is $-32E$.
\[ \boxed{-32E} \]
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