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# A single electron orbits around a stationary nucleus of charge +Ze where Z is atomic number and 'e' is the magnitude of the electric charge. The hydrogen like species required 47.2eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find(i) the value of Z and give the hydrogen like species formed.(ii) the kinetic energy and potential energy of the electron in the first Bohr orbit.

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Solution

## The above is the formula for the change in energy for a general hydrogen-like atom.Here, E0 is the ionisation energy of the hydrogen atom.Now, ΔE = Z2×(13.6 122 - 132)= 47.2= Z2× 13.636 ×5 = 47. 2= Z2= 47.2×3613.6×5 = 25Thus the value of Z is Z = 5Now, the kinetic energy of first Bohr orbit is equal to the energy of the orbit.So, E1 = −Z2E0= -25 × 13.6 eVTherefore,K.E= 544×10−19 JPotential energy of the electron = −2×K.E = −2× 544×10−19 J= −1088×10−19 J   Suggest Corrections  0      Similar questions  Related Videos   Emission and Absorption Spectra
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