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:

• A. $-32E$
• B. $-8E$
• C. $-4E$
• D. $-16E$

Hint:

For any Bohr orbit, $PE = -2 KE$.

Answer 1

The Correct Option is A

Step 1: Find the orbit number.
Bohr's rule says the angular momentum of an electron is $L = \dfrac{nh}{2\pi}$. We are given $L = \dfrac{2h}{\pi} = \dfrac{4h}{2\pi}$, so matching gives $n = 4$.

Step 2: Recall the kinetic energy formula.
For hydrogen the kinetic energy in orbit $n$ is $KE_n = \dfrac{13.6}{n^2}\ eV$.

Step 3: Apply it to the fourth orbit.
In the $n = 4$ orbit, $KE_4 = \dfrac{13.6}{16}\ eV$. The problem calls this $E$, so $E = \dfrac{13.6}{16}$, which means $13.6 = 16E$.

Step 4: Find the kinetic energy of the first orbit.
For $n = 1$, $KE_1 = \dfrac{13.6}{1} = 13.6\ eV$. Using $13.6 = 16E$, this is $KE_1 = 16E$.

Step 5: Use the energy relation.
In any Bohr orbit the potential energy is twice the kinetic energy in size but negative: $PE = -2\,KE$.

Step 6: Get the potential energy of the first orbit.
So $PE_1 = -2 \times 16E = -32E$. \[ \boxed{-32E} \]