A monochromatic light source of intensity 5 mW emits 8 $\times 10^{15}$ photons per second. This light ejects photo electrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

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Solution

Given

$W_{0} = h v - e v_{0}$

$= \frac{5 \times 10^{- 3}}{8 \times 10^{15}} - 1.6 \times 10^{- 19} \times 2$

$= 6.25 \times 10^{- 19} - 3.2 \times 10^{- 19}$

$= 3.05 \times 10^{- 19}$

$= \frac{3.05 \times 10^{- 19}}{1.6 \times 10^{- 15}} = 1.906 e V .$

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A monochromatic light source of intensity 5 mW emits 8 ×1015 photons per second. This light ejects photo electrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Given W0=hv−ev0 =5×10−38×1015−1.6×10−19×2 =6.25×10−19−3.2×10−19 =3.05×10−19 =3.05×10−191.6×10−15=1.906 eV.

A monochromatic light source of intensity 5 mW emits 8 $\times 10^{15}$ photons per second. This light ejects photo electrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Given

$W_{0} = h v - e v_{0}$

$= \frac{5 \times 10^{- 3}}{8 \times 10^{15}} - 1.6 \times 10^{- 19} \times 2$

$= 6.25 \times 10^{- 19} - 3.2 \times 10^{- 19}$

$= 3.05 \times 10^{- 19}$

$= \frac{3.05 \times 10^{- 19}}{1.6 \times 10^{- 15}} = 1.906 e V .$