A μ-meson particle of charge equal to that of an electron, -e and mass 208 times the mass of the electron moves in a circular orbit around the nucleus of charge +4e. Assuming the bohr's model of the atom to be applicable to the system, the inverse of wavelength of the photon emitted when an electron jumps in the 4rth orbit of the atom from infinity in terms of rydberg constant is given by βRh. find the value of β/26. (Rh-Rhydberg's constant)
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Mass of the particle = 208 m
Inverse of Wavelength refers to the Wavenumber.
Wave number = Rydbergs Constant × Z² [ 1 / n₁² - 1 / n₂² ]
Given that Charge is +4e. Therefore number of protons is 4. Therefore atomic number of the given particle is 4.
⇒ Wave Number = Rydbergs Constant × 4² [ 1 / 4² - 1 / ∞² ]
⇒ Wave Number = Rydbergs Constant × 16 [ 1/16 - 0 ]
⇒ Wave Number = Rydbergs Constant × 1 = Rh
Therefore β in the above case is 1.
Therefore value of β/26 is 1/26.
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