To find the wavelength of radiation emitted due to the transition of an electron from the n=4 state to the n=2 state in a hydrogen atom, we can follow these steps:
### Step 1: Understand the Energy Levels
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.
### Step 2: Calculate the Energy for n=4 and n=2
We need to calculate the energies for \( n=4 \) and \( n=2 \):
- For \( n=4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]
- For \( n=2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
### Step 3: Calculate the Change in Energy
The change in energy (\( \Delta E \)) during the transition from \( n=4 \) to \( n=2 \) is:
\[ \Delta E = E_2 - E_4 = (-3.4 \, \text{eV}) - (-0.85 \, \text{eV}) \]
\[ \Delta E = -3.4 + 0.85 = -2.55 \, \text{eV} \]
### Step 4: Relate Energy to Wavelength
The energy of the emitted photon can also be related to its wavelength (\( \lambda \)) using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
### Step 5: Solve for Wavelength
Rearranging the equation gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2.55 \, \text{eV}} \]
### Step 6: Convert Energy from eV to Joules
Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = 2.55 \, \text{eV} = 2.55 \times 1.6 \times 10^{-19} \, \text{J} = 4.08 \times 10^{-19} \, \text{J} \]
### Step 7: Calculate Wavelength
Now substituting \( E \) in Joules into the wavelength formula:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4.08 \times 10^{-19}} \]
Calculating this gives:
\[ \lambda \approx 4.86 \times 10^{-7} \, \text{m} = 486 \, \text{nm} \]
### Final Answer:
The wavelength of radiation emitted due to the transition from \( n=4 \) to \( n=2 \) in a hydrogen atom is approximately \( 486 \, \text{nm} \).
---
