As the electron in the Bohr orbit is hydrogen atom passes from state `n = 2` to `n = 1` , the `KE (K) and PE (U)` charge as

Options

A. K two-fold. U also two-fold
B. K four-fold. U also four-fold
C. K four-fold. U also two-fold
D. K two-fold. U also four-fold

Explanation

To solve the problem of how the kinetic energy (KE) and potential energy (PE) change as the electron in a hydrogen atom transitions from the state \( n = 2 \) to \( n = 1 \), we can follow these steps:

### Step 1: Understand the relationship of KE and PE in Bohr's model
In Bohr's model of the hydrogen atom:
- The kinetic energy (KE) of the electron in the nth orbit is given by:
\[ KE = -\frac{1}{2} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2 n^2} \]
where \( Z \) is the atomic number (1 for hydrogen), \( e \) is the charge of the electron, \( m \) is the mass of the electron, and \( n \) is the principal quantum number.

- The potential energy (PE) of the electron is given by:
\[ PE = -\frac{Z e^2}{4 \pi \epsilon_0 r} \]
where \( r \) is the radius of the orbit, which is proportional to \( n^2 \).

### Step 2: Calculate KE and PE for \( n = 2 \) and \( n = 1 \)
1. For \( n = 2 \):
- \( KE_2 = -\frac{1}{2} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2 (2^2)} = -\frac{1}{8} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2} \)
- \( PE_2 = -\frac{Z e^2}{4 \pi \epsilon_0 (2^2)} = -\frac{1}{4} \frac{Z e^2}{4 \pi \epsilon_0} \)

2. For \( n = 1 \):
- \( KE_1 = -\frac{1}{2} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2 (1^2)} = -\frac{1}{2} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2} \)
- \( PE_1 = -\frac{Z e^2}{4 \pi \epsilon_0 (1^2)} = -\frac{Z e^2}{4 \pi \epsilon_0} \)

### Step 3: Compare KE and PE between the two states
- The ratio of kinetic energies:
\[ \frac{KE_1}{KE_2} = \frac{-\frac{1}{2} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2}}{-\frac{1}{8} \frac{Z^2 e^4 m}{(4 \pi \epsilon_0)^2}} = 4 \]
Thus, \( KE_1 = 4 \times KE_2 \).

- The ratio of potential energies:
\[ \frac{PE_1}{PE_2} = \frac{-\frac{Z e^2}{4 \pi \epsilon_0}}{-\frac{1}{4} \frac{Z e^2}{4 \pi \epsilon_0}} = 4 \]
Thus, \( PE_1 = 4 \times PE_2 \).

### Conclusion
As the electron transitions from \( n = 2 \) to \( n = 1 \):
- The kinetic energy increases by a factor of 4.
- The potential energy also increases by a factor of 4.

### Final Answer
- \( KE \) changes to \( 4 \times KE_2 \)
- \( PE \) changes to \( 4 \times PE_2 \)