The threshold frequency of cesium is 5.16 tim | vedclass

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Question

(B) The work function $\phi_{0}$ is given by the formula $\phi_{0} = h 
u_{0}$,where $h$ is Planck's constant $(6.626 	imes 10^{-34} \ J \cdot s)$ a

Vedclass · Accepted Answer

$2.14$

Vedclass · Answer

(B) The work function $\phi_{0}$ is given by the formula $\phi_{0} = h 
u_{0}$,where $h$ is Planck's constant $(6.626 	imes 10^{-34} \ J \cdot s)$ and $
u_{0}$ is the threshold frequency.To convert the energy from Joules to electron-volts $(eV)$,we divide by the charge of an electron $(e = 1.6 	imes 10^{-19} \ C)$.$\phi_{0} = \frac{h 
u_{0}}{e} = \frac{6.626 	imes 10^{-34} 	imes 5.16 	imes 10^{14}}{1.6 	imes 10^{-19}} \ eV$.$\phi_{0} \approx \frac{34.19 	imes 10^{-20}}{1.6 	imes 10^{-19}} \ eV$.$\phi_{0} \approx 2.137 \ eV$.Rounding to two decimal places,we get $\phi_{0} \approx 2.14 \ eV$.

The threshold frequency of cesium is $5.16 \times 10^{14} \ Hz$. Then its work function is . . . . . . $eV$.

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