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Ans 2.(i)E=hc/
 substitute the values ans is 4.97 *10-19 (ii)K.E=hv-hv 0substitute the values ans is .97 eV or 1.56*10-19 J (III) V = (K.E *2/9.1*10-31)1/2=5.85*105 m/s . Ans 3.I.E of Na =E of irradiated photon=hc/*N=495kJ/mole. Ans.4 E of 1 photon=hc/=3.49*10-19 J ; 25 watt=25 J/s=E of bulb; Number of photons emitted=25/3.49 *10-19=7.16 *1019 where N=6.023 *1023 |
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