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Calculate and compare the energies of two radiations one with wavelength 800 pm and the other with wavelength 400 pm. |
Answer» `E=hv=(hc)/(lambda)` `"First case": " "lambda=800"pm"=800xx10^(-12)=8xx10^(-10)m` `c=3.0xx10^(8)m s^(-1), h=6.626xx10^(-34)Js` `:. " "E_(1)=((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((4xx10^(-10)m))=4.96xx10^(-16)J` `"Second case": " "lambda=400"pm"=400xx10^(-12)=4xx10^(-10)m` `" "c=3.0xx10^(8)m s^(-1), h=6.626xx10^(-34)Js` `:. " "E_(2)=((6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((4xx10^(-10)m))=4.96xx10^(-16)J` `" " (E_(1))/(E_(2))=(2.48xx10^(-16)J)/(4.96xx10^(-16)J)=(1)/(2) or E_(2)=2E_(1)`. |
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