Question

A hydrogen sample is prepared in a particular excited state. Photons of energy 2.55 V got absorbed
into the sample to take some of the electrons to a further excited state B. Find orbit numbers of the
states A and B. Given the allowed energies of hydrogen atom:
E 1 3.6 eV, E.-3.4 V, E = -1.5 V, E4--0.85 eV. Es - -0.54 ev​

Answer 1

Answer:

2 and 4.

Explanation:

After observing all the energies we will get that the transmission of the electrons will take place from the -3.4 and -0.85 to make energy of 2.55 volt to get absorbed into the sample.

So, we get that the orbit or the quantum number of the states of A will be 4 since the energy is -3.4 and the quantum number of the B will be 2 since the energy is -0.85. So, 2 and 4 are the orbital number of B and A respectively.

Answer 2

Answer:

By using energy formula, [ Given energy = 2.55eV]

hc/lambda = 13.6× Z^2 ( 1/n1^2 - 1/n2^2)

2.55=13.6(1^2) (1/n1^2 - 1/n2^2)

By solving, we get "n1 = 2", "n2 = 4".