The figure shows an energy level diagram for the hydrogen...

The figure shows an energy level diagram for the hydrogen atom. Several transitions are shown and are labeled by letters. In which transition is a Balmer series photon absorbed? Which transition corresponds to the absorption of the photon with the longest wavelength? Determine the energy of the photon involved in transition E Determine the wavelength of the radiation involved in transition B Which transition will occur when a hydrogen atom is irradiated with radiation of wavelength 103 nm? Which transition will occur when a hydrogen atom is irradiated with radiation of frequency 1.60 x 10 14 Hz? Principal Quantum Number ? 4 3 2 1 Note: The diagram is not drawn to scale.

Transcript

00:01 Hello, the question is taken from physics and the question is the figure shows an energy level diagram of the hydrogen atom.

00:07 Several transitions are shown and are labeled by later.

00:11 In which transition is a balmer series photon absorbed? in which transition is a balmer series photon absorbed? so balmer series is basically corresponding to the second one.

00:23 So balmer series photon is absorbed in corresponding to the e transition in which a photon photon is absorbed in the second state absorbed by the second state atom and move to the third so the answer is option three so what happened in case of third this is second and this is third a photon will come here so it is absorbed and the atom move to the third third excited state okay and so second excited state next is what's what transition corresponding to absorption of photon of longest wavelength okay so wavelength will be longest when energy difference is small so longest wavelength is corresponding to c okay so when delta e is small wavelength will be that is the case of c and this is third parties determine the energy of photon involved in transition e so delta e is equal to minus 13 .6 over to 3 square plus 13 point there is a minor sign between them but due to the negative signs of this become 13 .6 over 2 square in electron volt so that will be let me evaluate this value 1 .89 electron volt okay and fourth part is determine the wavelength of the radiation involved in transition b e4 to e2 so delta e is equal to 13 .6 divided by minus 4 to 2 plus 13 .6 divided by 4 square eva so that will be minus 2 .55 electron volt okay and fifth part is what transition will occur when a hydrogen atom is irradiated with a radiation of a wavelength 103 nanometer so when lambda is equal to 103 nanometer then what is the value of delta e basically what is the energy difference involved so so from here delta e is equal to hc by lambda so that will be h is equal to 6 .62 into 10 to the power minus 34 into c is 3 10 to the power 8 wavelength is 103 into 10 to the power 9 wavelength is 10 to the power 9 into 1 .6 into 10 to the power minus 19 so this will give you the energy difference is equal to 6 .6 to e minus 34 multiply 3e divided by 103e minus 9 divided by 1 .6e minus 19 so that is when the energy difference is i did anything more one second i convert it into electron 1 .6 to e minus 34 multiply 38 multiply 103e minus 9 divided sorry divided by 1 .6e minus 19...

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