Dear Student, The computation is expressed below, KEKE'=hv-hvohv'-hvo=v-vov'-voKE2KE'=2 0×1016-vo3 2×1016-vo0 5(3 2×1016-vo)=2 0×1016-vo1 6×1016-0 5vo=2 0×1016-vovo-0 5v0=(2 0-1 6)×10160 5vo=0 4×1016vo=0 8×1016vo=8×1015 HzHence the answer is Option (2) vo=8×1015 Hz Regards

Ans question no 14

Ans question no 14 A certain metal when irradiated with light 22. (v • 32 Hz) omits photo electrons with kinetic, energy as did photo electrorv when the same metal is irradiated by light 'v 20 10 to Hz). Calculate of electro)? v. (3) 12 10'6 Hz —313 6/n7 kcal/mole. If the value Of E -34.84 kcal/mole, to which value does •n' he rnoving With 'The mom BroglE (1) 3.2 (3) 5.3 x The u in the n should correspond? (2jzJ WHICh transition of with same energy change as n = 6 to n = 4 transition -ih (4) n = 2 ton = 1 Zeeman effect refers to the (1) Splitting of the spectral lines in a magnetic (2) Splitting up of the spectral lines in an electrostatic field 24. 25. The s electr (3) 26

Dear Student,

The computation is expressed below,

\frac{K E}{K E^{'}} = \frac{h v - h v_{o}}{h v^{'} - h v_{o}} = \frac{v - v_{o}}{v^{'} - v_{o}} \frac{K E}{2 {K E}^{'}} = \frac{2.0 \times 10^{16} - v_{o}}{3.2 \times 10^{16} - v_{o}} 0.5 (3.2 \times 10^{16} - v_{o}) = 2.0 \times 10^{16} - v_{o} 1.6 \times 10^{16} - 0.5 v_{o} = 2.0 \times 10^{16} - v_{o} v_{o} - 0.5 v_{0} = (2.0 - 1.6) \times 10^{16} 0.5 v_{o} = 0.4 \times 10^{16} v_{o} = 0.8 \times 10^{16} v_{o} = 8 \times 10^{15} H z H e n c e t h e a n s w e r i s O p t i o n (2) v_{o} = 8 \times 10^{15} H z

Regards.

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