When a photon is emitted from an atom , the atom recils The kinetic energy of recoils and the energy of the photon come from the difference in energy between the state involved in the transition suppose a hydrogen atom change its state from `n = 3 to n = 2` calculate the fractional change in the wavelength of light emitted , due to the recoil

Explanation

To solve the problem of calculating the fractional change in the wavelength of light emitted due to the recoil of a hydrogen atom transitioning from \( n = 3 \) to \( n = 2 \), we will follow these steps:

### Step 1: Calculate the energy difference (\( \Delta E \)) between the two states
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]
For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]
Now, calculate the energy difference:
\[ \Delta E = E_2 - E_3 = -3.4 - (-1.51) = -3.4 + 1.51 = -1.89 \, \text{eV} \]
Thus, the energy difference is approximately \( \Delta E \approx 1.89 \, \text{eV} \).

### Step 2: Calculate the wavelength (\( \lambda \)) of the emitted photon
Using the energy-wavelength relationship:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{\Delta E} \]
Where:
- \( h \) (Planck's constant) \( \approx 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) \( \approx 3 \times 10^8 \, \text{m/s} \)
- Convert \( \Delta E \) from eV to Joules: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)
\[ \Delta E \approx 1.89 \times 1.6 \times 10^{-19} \approx 3.024 \times 10^{-19} \, \text{J} \]
Now, substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3.024 \times 10^{-19}} \approx 6.58 \times 10^{-7} \, \text{m} \approx 658 \, \text{nm} \]

### Step 3: Consider the recoil of the atom
When the photon is emitted, the atom recoils. The momentum conservation gives:
\[ 0 = -mv + \frac{h}{\lambda'} \]
Where \( \lambda' \) is the wavelength after recoil. The kinetic energy of the recoiling atom is:
\[ KE = \frac{1}{2} mv^2 \]
Using the energy conservation:
\[ \Delta E = \frac{hc}{\lambda'} + \frac{1}{2} mv^2 \]
Substituting \( mv = \frac{h}{\lambda'} \) into the kinetic energy equation, we can express \( v \) in terms of \( \lambda' \):
\[ KE = \frac{1}{2} m \left(\frac{h}{\lambda'}\right)^2 \]

### Step 4: Relate \( \lambda' \) to \( \lambda \)
Using the approximation for small changes in wavelength due to recoil:
\[ \lambda' = \lambda \left(1 + \frac{\Delta \lambda}{\lambda}\right) \]
Where \( \Delta \lambda \) is the change in wavelength due to recoil.

### Step 5: Calculate the fractional change in wavelength
The fractional change in wavelength is given by:
\[ \frac{\Delta \lambda}{\lambda} = \frac{2 \Delta E}{mc^2} \]
Where \( m \) is the mass of the hydrogen atom \( (m \approx 1.67 \times 10^{-27} \, \text{kg}) \) and \( c \) is the speed of light.

Substituting the values:
\[ \frac{\Delta \lambda}{\lambda} = \frac{2 \times 1.89 \times 1.6 \times 10^{-19}}{(1.67 \times 10^{-27})(3 \times 10^8)^2} \]
Calculating this gives:
\[ \frac{\Delta \lambda}{\lambda} \approx 10^{-9} \]

### Final Answer
The fractional change in the wavelength of light emitted due to recoil is approximately \( 10^{-9} \).

---