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If the energy of a photon of light is given by E=kh^(x)c^(y)lambda^(z), where h= Planck's constant, c = velocity of light and lambda = wavelength then values of x,y,z in order are : |
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Answer» `1,2,1` Putting the dimensions `ML^(2)T^(-2)=[ML^(2)T^(-1)]^(x)[L^(1)T^(-1)]^(y)[L^(1)]^(z)` `[ML^(2)T^(-2)]=M^(x)L^(2x+y+z)T^(-x-y)` `:.`COMPARING the dimensions `x=1|{:(2x+y+z=1),(2+1+z=2),(z=-1):}||{:(-x-y=-2),(-1-y=-2),(y=1):}|` `:.`Correct choice is `(d)`. |
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