When an electron in the hydrogen atom makes a transition from an orbital with n=5 to an orbital with n=3, what is the wavelength (in nm) of the light emitted from the hydrogen atom? Round and report your answer to the nearest whole number. Don’t enter the unit in the answer box. C=3.0x10^8 m/s, h=6.626x10^-34 JxS

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Chapter7: Quantum Theory Of The Atom
Section: Chapter Questions
Problem 7.52QP: What is the wavelength of the electromagnetic radiation emitted from a hydrogen atom when the...
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When an electron in the hydrogen atom makes a transition from an orbital with n=5 to an orbital with n=3, what is the wavelength (in nm) of the light emitted from the hydrogen atom? Round and report your answer to the nearest whole number. Don’t enter the unit in the answer box. C=3.0x10^8 m/s, h=6.626x10^-34 JxS
Expert Solution
Step 1: Determination of energy released in this transition

Introduction:

Given that electron is moving from n=5 to n=3. When the transition is taking from a higher energy level to a lower energy level it is called as deexcitation. Energy is released during electronic deexcitation.

The energy released in deexcitation is equal to the difference in energies of initial and final levels. Negative sign is used to indicate that the energy is released in this process.

The energy released for electronic deexcitation is given by :

                 Ef-Ei=-2.18×10-18J(1nf2-1ni2)Substituting the corresponding values in the above equation gives:E=-2.18×10-18J×152-132E=-2.18×10-18J×125-19E=-2.18×10-18J×9-25225E=-2.18×10-18J×-16225E=1.55×10-19J

Therefore energy released during this transition is 1.55 x 10-19J. 

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