Question
A monochromatic light source of intensity $6mW$ emits $8 \times 10^{16}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is $2.0V$. Calculate the work function of the metal.
✓
Answer
$\text{w}_0=\text{hv}-\text{ev}_0$$=\frac{5\times10^{-3}}{8\times10^{15}}-1.6\times10^{-19}\times2$
(Given $V_0 = 2V$, No. of photons = $8 \times 10^{15}$, Power = 5mW)$=6.25\times10^{-19}-3.2\times10^{-19}=3.05\times10^{-19}\text{J}$
$=\frac{3.05\times10^{-19}}{1.6\times10^{-19}}=1.906\text{ev.}$
(Given $V_0 = 2V$, No. of photons = $8 \times 10^{15}$, Power = 5mW)$=6.25\times10^{-19}-3.2\times10^{-19}=3.05\times10^{-19}\text{J}$
$=\frac{3.05\times10^{-19}}{1.6\times10^{-19}}=1.906\text{ev.}$
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