First slide

Photoelectric effect and study of photoelectric effect

Question

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is : (Planck's constant $h = 6.62 \times 10^{- 34} J \cdot s$ )

Moderate

A

2.59 eV
B

2.10 eV
C

2.27 eV
D

1.93 eV

Solution

$h f = ϕ + e V_{s t o p}$
$ϕ$ is work function, which by definition is the minimum energy for the surface electron’s ejection.
Re-arranging, $V_{s t o p} = \frac{h}{e} f - \frac{ϕ}{e}$
To find $ϕ$ , point B is chosen because stopping potential is zero at this point.
x intercept $= \frac{ϕ}{h} = 5.5 \times 10^{14} H z$
$∴ ϕ = (6.62 \times 10^{- 34}) (5.5 \times 10^{14}) J$
$= \frac{(6.62 \times 10^{- 34}) (5.5 \times 10^{14})}{1.6 \times 10^{- 19}} e V$ $= 2.276 e V$

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