When the hydrogen atom is in its second excited state calculate the ratio of maximum to minimum wavelengths of radiation emitted in the process?

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Concept:

As per the Bohr model of an atom, an electron revolves around the nucleus in fixed orbits. While in these orbits it does not radiate energy. It radiates energy only when it jumps from one allowed energy level to another. Depending upon the difference in energy level, we get the electromagnetic spectrum of different wavelength that corresponds to different regions. 

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hcλEf−Ei=hν=hc

As per this formula, we compare the energy difference and find the wavelength of the transition.

For other series the table is shown below: 

.

Now looking at the table it's clear that the maximum wavelength emitted is emitted when the transition takes place from nf =1 to ni =2, 

the minimum wavelength emitted is emitted when the transition takes place from nf =1 to ni =  

Calculation:

In maximum wavelength, the electron falls from 1st excited state

In minimum wavelength, the electron comes from infinity

Hence, the ratio is 

The correct option is 1 

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Text Solution

Solution : In second excited state the electron is in `3^(rd)` orbit. To reach the ground state, the possible transitions will be n= 3 to n = 2,n =3 to n = 1, and n = 2 to n =1. <br> Thus the total number of spectral line observed will be three.

The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron, to ground state of Bohr's hydrogen atom is,

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