Fractional change in wavelength

[utkarshakash]

Homework Statement

When a photon is emitted from an atom the atom recoils. The kinetic energy of recoil and the energy of photon come from the difference in energies between the states involved in the transition. Suppose a hydrogen atom changes its state from n=3 to n=2. Calculate the fractional change in wavelength of light emitted due to the recoil.

Homework Equations

The Attempt at a Solution

Difference in energies of states = -13.6(1/4 - 1/9)

This is equal to sum of KE of recoil and energy of photon(ΔE).

$\dfrac{mv^2}{2} + \delta E = -13.6 \left( 1/4 - 1/9 \right)$
From this I can find energy of photon released only if I know the velocity v.

 

[haruspex]

What else might be conserved?

 

[utkarshakash]

What else might be conserved?

Momentum.
Using momentum conservation I can write

ΔE/c=mv.

Now if I plug v into energy conservation equation. I will get a quadratic in ΔE. Am I on the right track?

 

[haruspex]

Momentum.
Using momentum conservation I can write

ΔE/c=mv.

Now if I plug v into energy conservation equation. I will get a quadratic in ΔE. Am I on the right track?

That's what I would do.

 

[HalfLostAndSearching]

However, the velocity of recoil is not given in the question. To calculate the fractional change in wavelength, we need to know the initial and final wavelengths of the photon.

The fractional change in wavelength can be calculated as:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{\Delta E}{E}

Since the energy of the photon is directly proportional to its frequency and inversely proportional to its wavelength, the fractional change in wavelength can also be written as:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{\Delta \nu}{\nu} = \dfrac{\nu_f - \nu_i}{\nu_i}

Using the Rydberg formula, we can find the initial and final frequencies of the photon:

\nu_i = \dfrac{1}{\lambda_i} = R \left( \dfrac{1}{n_i^2} \right)

\nu_f = \dfrac{1}{\lambda_f} = R \left( \dfrac{1}{n_f^2} \right)

Substituting these values in the equation for the fractional change in wavelength, we get:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{R \left( \dfrac{1}{n_f^2} \right) - R \left( \dfrac{1}{n_i^2} \right)}{R \left( \dfrac{1}{n_i^2} \right)}

Simplifying this expression, we get:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{n_i^2 - n_f^2}{n_i^2}

Substituting the values n_i = 3 and n_f = 2, we get:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{3^2 - 2^2}{3^2} = \dfrac{5}{9}

Therefore, the fractional change in wavelength is 5/9 or approximately 0.56. This means that the wavelength of the emitted photon will decrease by approximately 56% due to the recoil of the hydrogen atom.