The kinetic energy of electrons emitted, when | vedclass

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(D) Given:Kinetic energy $(K.E.) = 2 	imes 10^{-19} \ J$Frequency of incident radiation $(
u) = 1.0 	imes 10^{15} \ Hz$Planck's constant $(h) = 6.6

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$6.97 	imes 10^{14}$

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(D) Given:Kinetic energy $(K.E.) = 2 	imes 10^{-19} \ J$Frequency of incident radiation $(
u) = 1.0 	imes 10^{15} \ Hz$Planck's constant $(h) = 6.6 	imes 10^{-34} \ Js$According to Einstein's photoelectric equation:$K.E. = h(
u - 
u_0)$Where $
u_0$ is the threshold frequency.Substituting the values:$2 	imes 10^{-19} = 6.6 	imes 10^{-34} 	imes (1.0 	imes 10^{15} - 
u_0)$$\frac{2 	imes 10^{-19}}{6.6 	imes 10^{-34}} = 1.0 	imes 10^{15} - 
u_0$$0.303 	imes 10^{15} = 1.0 	imes 10^{15} - 
u_0$$
u_0 = 1.0 	imes 10^{15} - 0.303 	imes 10^{15}$$
u_0 = 0.697 	imes 10^{15} \ Hz = 6.97 	imes 10^{14} \ Hz$

The kinetic energy of electrons emitted, when radiation of frequency $1.0 \times 10^{15} \ Hz$ hits a metal, is $2 \times 10^{-19} \ J$. What is the threshold frequency of the metal (in $Hz$)?
$(h = 6.6 \times 10^{-34} \ Js)$

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The kinetic energy of electrons emitted, when radiation of frequency $1.0 \times 10^{15} \ Hz$ hits a metal, is $2 \times 10^{-19} \ J$. What is the threshold frequency of the metal (in $Hz$)?$(h = 6.6 \times 10^{-34} \ Js)$