A `pi-meason` hydrogen atom is a bound state of negative charged pion (denoted by `pi^(bar), m_(pi) = 273 m_(e))` and a proton. Estimate the number of revolutions a `pi-meason` makes (averagely ) in the ground state on the atom before , it decays (mean life of a `pi-meason ~= 10^(-8) s`, mass of proton `= 1.67 xx 10^(-27) kg)`.

Answer

`1.77 xx 10^(10)` revolutions

Explanation

The frequency of revolution of an electron in hydrogen atom in first orbit is given by
`f_(1) = (4 pi^(2) K^(2) e^(4) m)/(h^(2))`
Here, in case of a `pi-meson` the mass of electron in replied by the reduced mass as
`mu = (m_(p) (273) m)/(m_(p) + 273 m)`
Thus the frequency of revolution of `pi-meson` will become
`f_(1) = (4 pi^(2) Ke^(4)m_(p) (273) m)/(h^(2)(m_(p) + 273 m))`
`4 xx (3.14 xx 9 xx 10^(9) (1.6 xx 10^(-19))^(4)`
`= (xx (1.67 xx 10^(-27)) (273 xx 9.1 xx 10^(-31)))/((6.63 xx 10^(-34)) ^(3) (1.67 xx 10^(-27) + 273 xx 9.1 xx 10^(-31)))`
`= 1.77 xx 10^(18) s^(-1)`
Thus, in the lift of `10^(-8) s`, number of revolation made by the `pi-meson` is given as
`N = f_(1) xx Delta t`
or `= 1.77 xx 10^(18) xx 10^(-8)`
or `= 1.77 xx 10^(10)` revolutions