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The frequency of emission line for eny ...

The frequency of emission line for eny transition in positronium atom (consisting of a positron and an electron) is `x` times the frequency of the corresponding line in the case of `H` atom , where `x` is

A

`sqrt2`

B

`1//2 sqrt2`

C

`1//2 sqrt2`

D

`1//2`

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The correct Answer is:
To solve the problem of finding the ratio of the frequency of emission lines for transitions in a positronium atom compared to a hydrogen atom, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Energy Levels**: - The energy levels for positronium (E_n) can be expressed as: \[ E_n = -\frac{6.8}{n^2} \text{ eV} \] - The energy levels for hydrogen (E_n) can be expressed as: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \] 2. **Calculate the Energy Difference**: - For positronium, the energy difference between the second and first energy levels (E_2 - E_1) is: \[ E_{2p} - E_{1p} = E_2 - E_1 = -\frac{6.8}{2^2} - \left(-\frac{6.8}{1^2}\right) = -\frac{6.8}{4} + 6.8 = 6.8 \left(1 - \frac{1}{4}\right) = 6.8 \cdot \frac{3}{4} \] - For hydrogen, the energy difference between the second and first energy levels (E_2 - E_1) is: \[ E_{2h} - E_{1h} = E_2 - E_1 = -\frac{13.6}{2^2} - \left(-\frac{13.6}{1^2}\right) = -\frac{13.6}{4} + 13.6 = 13.6 \left(1 - \frac{1}{4}\right) = 13.6 \cdot \frac{3}{4} \] 3. **Relate Energy Difference to Frequency**: - The frequency of the emitted photon (ν) is related to the energy difference by: \[ E = h \nu \] - Thus, for positronium: \[ h \nu_p = 6.8 \cdot \frac{3}{4} \] - And for hydrogen: \[ h \nu_h = 13.6 \cdot \frac{3}{4} \] 4. **Find the Ratio of Frequencies**: - Now, we can find the ratio of the frequencies: \[ \frac{\nu_p}{\nu_h} = \frac{6.8 \cdot \frac{3}{4}}{13.6 \cdot \frac{3}{4}} = \frac{6.8}{13.6} \] - Simplifying this gives: \[ \frac{\nu_p}{\nu_h} = \frac{6.8}{13.6} = \frac{1}{2} \] 5. **Conclusion**: - Therefore, the value of \( x \) is: \[ x = \frac{1}{2} \] ### Final Answer: The frequency of the emission line for any transition in the positronium atom is \( \frac{1}{2} \) times the frequency of the corresponding line in the case of the hydrogen atom. ---

To solve the problem of finding the ratio of the frequency of emission lines for transitions in a positronium atom compared to a hydrogen atom, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Energy Levels**: - The energy levels for positronium (E_n) can be expressed as: \[ E_n = -\frac{6.8}{n^2} \text{ eV} ...
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