The ground state of an unidentified hydrogen like ion is - 217.6ev. calculate the atomic no and identify the?
Answer:
The energy of an electron in a hydrogen-like atom or ion is given by:
E(n) = -2*(pi^2 *Z^2 * u * e^4)/(h^2 * n^2)
where Z is the charge on the nucleus (the atomic number)
e is the value of the electronic charge = 4.803*10^-10 (cm^(3/2) * gm^1/2)/s
h is Planck's constant = 6.6261 * 10^-34 kg*m^2 / s
u is the reduced mass of the electron-nucleus system = (mass nucleus)*(mass electron)/(mass nucleus + mass) electron) ~= mass of electron = 9.1094*10^-31 kg
and
n is the principal quantum number.
Solving for Z, we have:
Z = sqrt((E * h^2 * n^2)/(-2 * pi^2 * m * e^4))
where I've let u = m = mass of electron.
In this case, we have that E = 217.6 eV = 3.486*10^-17 J, and n = 1.
If you plug in the appropriate values (watching your units!), you will find that Z = 4. This is a beryllium ion.
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