When an electron jumps from `n_(1) th orbit to `n_(2)` th orbit, the energy radiated is given by

Options

A. `h v = E_(1)//E_(2)`
B. `h v = E_(2)//E_(1)`
C. `h v = E_(1) - E_(2)`
D. `h v = E_(2) - E_(1)`

Explanation

To solve the problem of finding the energy radiated when an electron jumps from the \( n_1 \)th orbit to the \( n_2 \)th orbit, we can follow these steps:

### Step 1: Understand the Energy Levels
In a hydrogen-like atom, the energy of an electron in the \( n \)th orbit is given by the formula:
\[ E_n = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n^2} \]
where:
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( k \) is Coulomb's constant,
- \( e \) is the charge of the electron,
- \( m \) is the mass of the electron,
- \( \hbar \) is the reduced Planck's constant,
- \( n \) is the principal quantum number.

### Step 2: Calculate the Energy in Both Orbits
For the \( n_1 \)th orbit, the energy is:
\[ E_{n_1} = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n_1^2} \]
For the \( n_2 \)th orbit, the energy is:
\[ E_{n_2} = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n_2^2} \]

### Step 3: Determine the Energy Difference
When the electron transitions from the \( n_1 \)th orbit to the \( n_2 \)th orbit, the energy radiated (or emitted) is the difference in energy between these two states:
\[ E_{\text{radiated}} = E_{n_1} - E_{n_2} \]

Substituting the expressions for \( E_{n_1} \) and \( E_{n_2} \):
\[ E_{\text{radiated}} = \left(-\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n_1^2}\right) - \left(-\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n_2^2}\right) \]
\[ E_{\text{radiated}} = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2} \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]

### Step 4: Final Expression for Energy Radiated
Thus, the energy radiated when an electron jumps from the \( n_1 \)th orbit to the \( n_2 \)th orbit is given by:
\[ E_{\text{radiated}} = \frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2} \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]