Question

The strongest lines of the He spectrum occur at the following wavelengths ; 0) 127. 57 nm, (2) 164./Z nm, and (3) 25.74 nm. Calculate the transitions associated with these lines

Answer 1

Rydberg equation to fine out wavelength:

1/λ = R_­H­. Z² (1/n­₁­² - 1/n­₂­²)

where λ = wavelength of the emitted radiation in cm

              R_­H­ = Rydberg's constant = 109677 cm^-1

              n_­1­ & n­₂­ = energy levels (always n2 > n1)

            here n1 = 1 and n2 = 5

            Z = atomic number (for He⁺ , Z=2)

Wavelength Relationship 1/1 = RH. ZZ (1/n- 12-1/n2?) 1/n2-1/nz? 127.57 nm 0.1786 1/(127.57 X 107) = 109677*4*(1/nı- 1/n2?) 164.12 nm 0.1389 1/164.57 X 10 ) = 109677*4*(1/n? - 1/n22) 25.64 nm 0.8890 1/(25.64 X 107) = 109677*4*(1/n,? - 1/n2?)

Wavelength Problem 1C (1/n2-1/n- Trial and error method Transition 127.57 nm 0.1786 n=4 ------>n=2 ni = 2 & n2 =4 approximately equal to 0.1786 164.12 nm 0.1389 n=3 ------>n=2 na = 2 & n2 =3 approximately equal to 0.1389 25.64 nm 0.8890 n=3 ------>n=1 ni = 1 & n2 =3 approximately equal to 0.8890

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