a hydrogen atom jumps from third excited state to first excited state the energy of photon emitted in the process is
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A hydrogen atom (ionisation energy=13.6) jumps from the third excited state to the first excited state. What is the energy of the photon emitted in the process?
For hydrogen atom in 3rd excited state, the energy is E_3=13.6/3^2 eV and similarly for hydrogen atom in 1st excited state, the energy is E_1=13.6/1^2 eV. So taking the difference in energies of two states (delta E) and equating it to ‘h*f’ should be helpful to get emitted photon energy which will be simply equal to ‘delta E’. In this case emitted photon energy is 12.1 eV.
For hydrogen atom in 3rd excited state, the energy is E_3=13.6/3^2 eV and similarly for hydrogen atom in 1st excited state, the energy is E_1=13.6/1^2 eV. So taking the difference in energies of two states (delta E) and equating it to ‘h*f’ should be helpful to get emitted photon energy which will be simply equal to ‘delta E’. In this case emitted photon energy is 12.1 eV.
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Answer:
12.09 eV
Explanation:
energy=13.6[1/(n1^2)-1/(n2^2)]
energy=13.6[1/(1^2)-1/(3^2)]
=13.6[(9-1)/9]
=13.6*(8/9)
=12.09 eV
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