Question

A hydrogen atom changes its state from \( n = 3 \) to \( n = 2 \). Due to recoil, the percentage change in the wavelength of emitted light is approximately \( 1 \times 10^{-n} \). The value of \( n \) is ______. \([ \text{Given: } Rhc = 13.6 \, \text{eV}, \, hc = 1242 \, \text{eV nm}, \, h = 6.6 \times 10^{-34} \, \text{J s}, \, \text{mass of the hydrogen atom} = 1.6 \times 10^{-27} \, \text{kg} ]\)

Answer 1

Correct Answer: 7

The change in energy during the transition is given by:
 \[ \Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 1.9 \, \text{eV} \]

The relationship between energy and wavelength is given by:
 \[ \Delta E = \frac{hc}{\lambda} \]

Rearranging for \( \lambda \):
 \[ \lambda = \frac{hc}{\Delta E} \]

To find the effect of recoil, consider the conservation of momentum:
 \[ P_i = P_f \] \[ 0 = -mv + \frac{h}{\lambda'} \]

Solving for the velocity \( v \) of the recoiling atom:
 \[ v = \frac{h}{m\lambda'} \]

The total energy change \( \Delta E' \) considering kinetic energy and emitted photon energy is given by:
 \[ \Delta E' = \frac{1}{2} mv^2 + \frac{hc}{\lambda'} \]

Substituting for \( v \) and simplifying:
 \[ \Delta E' = \frac{1}{2} \left( \frac{h}{m\lambda'} \right)^2 + \frac{hc}{\lambda'} \]

Using the energy conservation equation:
 \[ \lambda'^2 \Delta E - hc\lambda' - \frac{h^2}{2m} = 0 \]

Solving this quadratic equation for \( \lambda' \):
 \[ \lambda' = \frac{hc \pm \sqrt{h^2c^2 + 4\Delta E h^2 / 2m}}{2\Delta E} \]

Approximating for small changes:
 \[ \frac{\lambda' - \lambda}{\lambda} \approx \frac{\Delta E}{2mc^2} \]

Substituting the given values:
 \[ \frac{\lambda' - \lambda}{\lambda} = \frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2} \approx 10^{-7} \]

The percentage change in wavelength is approximately \( 10^{-7} \).

Answer 2

Step 1: Given data.
Transition in a hydrogen atom: \( n = 3 \rightarrow n = 2 \).
We are asked to find the percentage change in wavelength of emitted light due to recoil of the atom.
Given constants:
\( Rhc = 13.6 \, \text{eV} \)
\( hc = 1242 \, \text{eV nm} \)
\( h = 6.6 \times 10^{-34} \, \text{J s} \)
\( m_H = 1.6 \times 10^{-27} \, \text{kg} \)

Step 2: Find energy of emitted photon (without recoil).
For hydrogen, the energy difference between levels is:
\[ E = Rhc \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] \[ E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{5}{36} \right) = 1.8889 \, \text{eV} \]
Step 3: Calculate corresponding wavelength.
\[ \lambda = \frac{hc}{E} = \frac{1242}{1.8889} \approx 657.5 \, \text{nm} \]
Step 4: Recoil energy of the hydrogen atom.
When a photon of energy \( E = h\nu \) is emitted, the atom recoils with momentum \( p = \frac{E}{c} \).
The recoil energy of the atom is:
\[ E_r = \frac{p^2}{2m_H} = \frac{E^2}{2m_H c^2} \]
Step 5: Fractional (percentage) change in photon energy or wavelength.
Since recoil reduces photon energy slightly:
\[ \frac{\Delta E}{E} = \frac{E_r}{E} = \frac{E}{2m_H c^2} \] Thus, the fractional change in wavelength (approximately same magnitude) is:
\[ \frac{\Delta \lambda}{\lambda} \approx \frac{E}{2m_H c^2} \]
Step 6: Substituting values.
Convert \( E \) to joules:
\[ E = 1.8889 \, \text{eV} = 1.8889 \times 1.6 \times 10^{-19} = 3.02 \times 10^{-19} \, \text{J} \] \[ \frac{E}{2m_H c^2} = \frac{3.02 \times 10^{-19}}{2 \times 1.6 \times 10^{-27} \times (3 \times 10^8)^2} \] \[ = \frac{3.02 \times 10^{-19}}{2.88 \times 10^{-10}} \approx 1.05 \times 10^{-9} \] In percentage: \[ \text{Percentage change} = 1.05 \times 10^{-7} \times 100 = 1 \times 10^{-7} \]
Step 7: Final Answer.
\[ \boxed{n = 7} \]