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(C) The work function $\phi_0$ is given by the formula: $\phi_0 = h 
u_0 = \frac{hc}{\lambda_0}$.Given: $h = 6.6 	imes 10^{-34} \ J \cdot s$,$c = 3

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$2.5$

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(C) The work function $\phi_0$ is given by the formula: $\phi_0 = h 
u_0 = \frac{hc}{\lambda_0}$.Given: $h = 6.6 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda_0 = 5000 \ \mathring A = 5000 	imes 10^{-10} \ m = 5 	imes 10^{-7} \ m$.Substituting the values:$\phi_0 = \frac{(6.6 	imes 10^{-34}) 	imes (3 	imes 10^8)}{5 	imes 10^{-7}} \ J$.$\phi_0 = \frac{19.8 	imes 10^{-26}}{5 	imes 10^{-7}} \ J = 3.96 	imes 10^{-19} \ J$.To convert this into $eV$,divide by $1.6 	imes 10^{-19} \ J/eV$:$\phi_0 = \frac{3.96 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV = 2.475 \ eV$.Rounding to the nearest value,we get $\phi_0 \approx 2.5 \ eV$.

In the photoelectric effect,the threshold wavelength of sodium is $5000 \ \mathring A$. Find its work function in $eV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

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