If the work function of a metal is 6.63 times | vedclass

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(A) The work function $\phi$ is related to the threshold wavelength $\lambda$ by the equation: $\phi = \frac{hc}{\lambda}$.Given $\phi = 6.63 	imes 1

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$300$

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(A) The work function $\phi$ is related to the threshold wavelength $\lambda$ by the equation: $\phi = \frac{hc}{\lambda}$.Given $\phi = 6.63 	imes 10^{-19} \ J$,$h = 6.63 	imes 10^{-34} \ J \ s$,and $c = 3 	imes 10^{8} \ m \ s^{-1}$.Substituting the values: $6.63 	imes 10^{-19} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{\lambda}$.$\lambda = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{6.63 	imes 10^{-19}} \ m$.$\lambda = 3 	imes 10^{-7} \ m$.Converting to nanometers $(nm)$: $\lambda = 3 	imes 10^{-7} 	imes 10^{9} \ nm = 300 \ nm$.

If the work function of a metal is $6.63 \times 10^{-19} \ J$,the maximum wavelength of the photon required to remove a photoelectron from the metal is $.... \ nm$. (Nearest integer)
[Given : $h = 6.63 \times 10^{-34} \ J \ s$,and $c = 3 \times 10^{8} \ m \ s^{-1}$ ]

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If the work function of a metal is $6.63 \times 10^{-19} \ J$,the maximum wavelength of the photon required to remove a photoelectron from the metal is $.... \ nm$. (Nearest integer)[Given : $h = 6.63 \times 10^{-34} \ J \ s$,and $c = 3 \times 10^{8} \ m \ s^{-1}$ ]