Question

As the electron in Bohr is orbit of hydrogen atom Passes from state \(\mathrm{n}=2\) to \(\mathrm{n}=1\), the \(\mathrm{K} . \mathrm{E}\). and Potential energy changes as (A) Two fold, also two fold (B) four fold, two fold (C) four fold, also four fold (D) two fold, four fold

Short answer

As the electron in the hydrogen atom transitions from n = 2 to n = 1, both the kinetic energy and potential energy change by a factor of 4. Therefore, the correct answer is (C) four fold, also four fold.

Step-by-step solution

1. Recall the formula for kinetic and potential energy in Bohr's model.

In Bohr's model, the kinetic energy (KE) and potential energy (PE) of an electron in the hydrogen atom, at a particular energy level n, are given by: KE = \(\frac{KZe^2}{2\cdot a_0\cdot n^2}\) and PE = \(\frac{-KZe^2}{a_0\cdot n^2}\) Here, K is Coulomb's constant, Z is the atomic number (Z = 1 for hydrogen), e is the elementary charge, and \(a_0\) is the Bohr's radius.

2. Calculate the ratio of kinetic energies and potential energies at the two principal quantum numbers n = 1 and n = 2.

We will find the ratio of the kinetic energy at n = 2 to the kinetic energy at n = 1, and similarly for the potential energy. \(\frac{KE_2}{KE_1}\) = \(\frac{\frac{KZe^2}{2\cdot a_0\cdot 4}}{\frac{KZe^2}{2\cdot a_0}}\) = \(\frac{1}{4}\) \(\frac{PE_2}{PE_1}\) = \(\frac{\frac{-KZe^2}{a_0 \cdot 4}}{\frac{-KZe^2}{a_0}}\) = \(\frac{1}{4}\)

3. Analyze the changes in kinetic and potential energy.

From the ratios calculated in step 2, we can see that as the electron transitions from n = 2 to n = 1, both the kinetic energy and potential energy change four-fold (reduced by a factor of 4). So, the correct answer is: (C) four fold, also four fold.