The excitation energy of an electron from second orbit to third orbit of a hydrogen-like atomor ion with +Ze nuclear charge is 47.2eV . If the energy of H-atom in the lowest energy state is -13.6 eV , then the value of Z is
Question
The excitation energy of an electron from second orbit to third orbit of a hydrogen-like atomor ion with +Ze nuclear charge is 47.2eV . If the energy of H-atom in the lowest energy state is -13.6 eV , then the value of Z is
Solution
The energy of an electron in a hydrogen-like atom or ion is given by the formula:
E = -13.6 * Z^2 / n^2 eV
where Z is the nuclear charge and n is the principal quantum number (the number of the orbit).
The energy difference between the second and third orbits is given as 47.2 eV. We can set up an equation using the energy formula for the second and third orbits:
E3 - E2 = 47.2 eV
Substituting the energy formula into this equation gives:
(-13.6 * Z^2 / 3^2) - (-13.6 * Z^2 / 2^2) = 47.2
Solving this equation for Z gives:
Z^2 * (1/4 - 1/9) = -47.2 / -13.6
Z^2 * (5/36) = 3.47
Z^2 = 3.47 * 36/5
Z^2 = 25
Taking the square root of both sides gives:
Z = 5
So the nuclear charge Z is 5.
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