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An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers :
  • a)
     z = 1/x
  • b)
    x = 9/4 
  • c)
     y = 5/27
  • d)
    z = 5/27
Correct answer is option 'B'. Can you explain this answer?
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To understand why option 'B' is the wrong answer, let's analyze the given scenario step by step.

1. Electron transition from the second excited state to the first excited state:
- When an electron transitions from a higher energy level to a lower energy level within an atom, it releases energy in the form of a photon.
- The wavelength of the emitted photon can be calculated using the Rydberg formula: 1/λ = R_H * (1/n_f^2 - 1/n_i^2), where λ is the wavelength, R_H is the Rydberg constant for hydrogen, and n_f and n_i are the final and initial energy levels, respectively.
- In this case, the electron is transitioning from the second excited state (n_i = 3) to the first excited state (n_f = 2).
- The ratio of the wavelengths in the two cases is given as x, so we can write x = λ_2/λ_3.

2. Electron transition from the first excited state to the ground state:
- Similarly, when the electron transitions from the first excited state (n_i = 2) to the ground state (n_f = 1), it emits another photon.
- The ratio of the wavelengths in this case is given as x (same as in the previous case), so we can write x = λ_1/λ_2.

To find the ratio of the wavelengths x, we need to find λ_1, λ_2, and λ_3 in terms of x.

From the Rydberg formula:
1/λ_2 = R_H * (1/2^2 - 1/3^2)
1/λ_3 = R_H * (1/1^2 - 1/3^2)

Simplifying the equations:
1/λ_2 = R_H * (1/4 - 1/9) = 5R_H/36
1/λ_3 = R_H * (1 - 1/9) = 8R_H/9

Now, let's express λ_1 and λ_2 in terms of λ_3 using the given ratio x = λ_2/λ_3:
λ_2 = xλ_3
λ_1 = xλ_2 = x(xλ_3) = x^2λ_3

Therefore, λ_1 = x^2λ_3.

Now, let's find the ratio of momentum and energy for the two cases:

3. Momentum of the photons:
- The momentum of a photon can be calculated using the formula p = h/λ, where p is the momentum, h is the Planck's constant, and λ is the wavelength.
- The ratio of the momenta in the two cases is given as y, so we can write y = p_1/p_2.

Using the wavelength expressions derived earlier:
p_1 = h/λ_1 = h/(x^2λ_3)
p_2 = h/λ_2 = h/(xλ_3)

Therefore, y = p_1/p_2 = (h/(x^2λ_3))/(h/(xλ_3)) = x.

4. Energy of the photons:
- The energy of a photon can be calculated using the equation E = hc/λ,
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An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers :a)z = 1/xb)x = 9/4c)y = 5/27d)z = 5/27Correct answer is option 'B'. Can you explain this answer?
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