When an electron jumps from higher energy orbit to lower energy orbit, the difference in the energies in the two orbits is radiated as quantum (photon) of….

Options

A. `E = mc^(2)`
B. `E =h/v`ℎ
C. `E=(hc)/lamda`
D. `E=lamda/(hc)`

Answer

To solve the question regarding the energy emitted when an electron transitions from a higher energy orbit to a lower energy orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
- Electrons in an atom occupy specific energy levels or orbits. When an electron transitions from a higher energy level (E2) to a lower energy level (E1), it loses energy.

2. **Energy Difference**:
- The energy difference (ΔE) between the two levels is given by:
\[ \Delta E = E2 - E1 \]
- This energy difference is what will be emitted as a photon.

3. **Photon Emission**:
- The energy lost by the electron is emitted in the form of a photon. The energy of the emitted photon (E_photon) can be expressed as:
\[ E_{\text{photon}} = h \nu \]
- Here, \(h\) is Planck's constant and \(\nu\) (nu) is the frequency of the emitted photon.

4. **Relating Energy and Wavelength**:
- The frequency of the photon can also be related to its wavelength (\(\lambda\)) using the equation:
\[ c = \nu \lambda \]
- Where \(c\) is the speed of light. Rearranging this gives:
\[ \nu = \frac{c}{\lambda} \]

5. **Final Expression for Photon Energy**:
- Substituting the expression for frequency into the photon energy equation gives:
\[ E_{\text{photon}} = h \frac{c}{\lambda} \]
- This shows that the energy of the emitted photon is directly related to the wavelength of the emitted light.

6. **Conclusion**:
- Therefore, when an electron jumps from a higher energy orbit to a lower energy orbit, the difference in the energies of the two orbits is radiated as a quantum (photon) of energy given by:
\[ E_{\text{photon}} = \frac{hc}{\lambda} \]