1.

Find the velocity of an electron emitted by a metal whose threshold frequency is 2.25 xx 10^(14) sec^(-1) . When exposed to visible light of wavelength 5.0 xx 10^(-7) m.

Answer»

Solution :`phi = hv_(0)`
`= 6.6 xx 10^(-34) xx 2.25 xx 10^(14) = 0.928 eV`
Total ENERGY given by the visible light `= ( h C)/(lambda)`
` = ( 6.6 xx 10^(-34) xx 3 xx 10^(8))/( 5.0 xx 10^(-7)) = 2.47 eV `
K.E. = Total energy `E - phi = 2.45 - 0.928 = 1.547 eV `
K.E. = `( 1)/(2) mv^(2)`
`1.547 xx 1.6 xx 10^(-14) = ( 1)/( 2) xx 9 xx 10^(-31) xx v^(2)`
`v = sqrt(( 1.547 xx 1.6 xx 10^(-19) xx 2 )/( 9 xx 10^(-31))) 7.4 xx 10^(5) m//`SEC


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