Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 16136 by Tinkutara last updated on 18/Jun/17

Photoelectric emission is observed from  a surface when lights of frequency n_1   and n_2  incident. If the ratio of maximum  kinetic energy in two cases is K : 1  then (Assume n_1  > n_2 ) threshold  frequency is  (1) (K − 1) × (Kn_2  − n_1 )  (2) ((Kn_1  − n_2 )/(1 − K))  (3) ((K − 1)/(Kn_1  − n_2 ))  (4) ((Kn_2  − n_1 )/(K − 1))

$$\mathrm{Photoelectric}\:\mathrm{emission}\:\mathrm{is}\:\mathrm{observed}\:\mathrm{from} \\ $$ $$\mathrm{a}\:\mathrm{surface}\:\mathrm{when}\:\mathrm{lights}\:\mathrm{of}\:\mathrm{frequency}\:{n}_{\mathrm{1}} \\ $$ $$\mathrm{and}\:{n}_{\mathrm{2}} \:\mathrm{incident}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{maximum} \\ $$ $$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{two}\:\mathrm{cases}\:\mathrm{is}\:\mathrm{K}\::\:\mathrm{1} \\ $$ $$\mathrm{then}\:\left(\mathrm{Assume}\:{n}_{\mathrm{1}} \:>\:{n}_{\mathrm{2}} \right)\:\mathrm{threshold} \\ $$ $$\mathrm{frequency}\:\mathrm{is} \\ $$ $$\left(\mathrm{1}\right)\:\left(\mathrm{K}\:−\:\mathrm{1}\right)\:×\:\left(\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} \right) \\ $$ $$\left(\mathrm{2}\right)\:\frac{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} }{\mathrm{1}\:−\:\mathrm{K}} \\ $$ $$\left(\mathrm{3}\right)\:\frac{\mathrm{K}\:−\:\mathrm{1}}{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} } \\ $$ $$\left(\mathrm{4}\right)\:\frac{\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} }{\mathrm{K}\:−\:\mathrm{1}} \\ $$

Answered by ajfour last updated on 18/Jun/17

   hn_1 −hn_0 =(K.E.)_(max,1)      hn_2 −hn_0 =(K.E.)_(max,2)   so,   ((n_1 −n_0 )/(n_2 −n_0 )) = (K/1)    n_1 −n_0 =Kn_2 −Kn_0      n_0 = ((Kn_2 −n_1 )/(K−1)) .

$$\:\:\:{hn}_{\mathrm{1}} −{hn}_{\mathrm{0}} =\left({K}.{E}.\right)_{{max},\mathrm{1}} \\ $$ $$\:\:\:{hn}_{\mathrm{2}} −{hn}_{\mathrm{0}} =\left({K}.{E}.\right)_{{max},\mathrm{2}} \\ $$ $${so},\:\:\:\frac{{n}_{\mathrm{1}} −{n}_{\mathrm{0}} }{{n}_{\mathrm{2}} −{n}_{\mathrm{0}} }\:=\:\frac{{K}}{\mathrm{1}} \\ $$ $$\:\:{n}_{\mathrm{1}} −{n}_{\mathrm{0}} ={Kn}_{\mathrm{2}} −{Kn}_{\mathrm{0}} \\ $$ $$\:\:\:{n}_{\mathrm{0}} =\:\frac{{Kn}_{\mathrm{2}} −{n}_{\mathrm{1}} }{{K}−\mathrm{1}}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com