Aditi Chauhan
Last Activity: 10 Years ago
En of H = -21.76 *10-19/n2 J
∴ En of He+ = -21.76 *10-19/n2 * Z2 J
∴ E3 of He+ = -21.76 *10-19 *4/9 J
Hence energy equivalent to E3 must be supplied to remove the electron from 3rd orbit of He+. Wavelength corresponding to this energy can be determined by applying the relation.
E = hc/λ
or λ = hc/E = 6.625 *10-34 *3 */ 108 *9/21.76 *10-19 *4 = 2055 *10-10 m = 2055 Å