1.

7 The energy of the electron in the second and third Bohr'sorbits of the hydrogen atom is-5.42 x 10-12 erg and-2.41x10-12 erg respectively. Calculate the wavelength of theemitted light when the electron drops from the third to the(1981)second orbit.

Answer»

E3 – E2 = hv = hc/λ – 2.41 ´ 10–12 – (– 5.42 ´ 10–12) = 6.626 x 10-27 x 3 x 1010/ λ

∴ λ = 6.626 x 10-27 x 3 x 1010 / 3.01 x 10-12

= 6.604 x 10–5 cm = 6.604 x 10–5 x 108 = 6604 Å



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