1.

What is the number of proton of light with wavelength `300`nm that provide `2J` of energy ?

Answer» Let N proton of light provide `2J` energy so
`:. E =Nhv= (Nhc)/(lambda)`
or `N = (E lambda)/(hc)`
`= (2 xx 300 xx 10^(-9)m)/(6.262 xx 10^(-34) xx 3 xx 10^(6) ms^(-1))`
`= 3.01 xx 10^(18)` photons


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