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The cutoff frequency for a certain element is 1.22 x 1015 Hz. What is its work function in eV?
Hint: 1 eV 1.60 x 10-19 J
=[?] eV

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The Cutoff Frequency For A Certain Element Is 122 X 1015 Hz What Is Its Work Function In EV Hint 1 EV 160 X 1019 J EV Someone Please Teach Me How To Solve This class=

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pstnonsonjoku

The work function of the metal is obtained as 5.1 eV

What is the cutoff frequency?

The cutoff frequency is the frequency below which photo electric effect can not occur as electrons are not removed from the metal surface.

Now;

fo = 1.22 x 10^15 Hz

Wo = hfo

Wo = 6.6 * 10^-34 * 1.22 x 10^15

Wo = 8.1 * 10^-19 J

If 1 eV = 1.60 x 10-19 J

x eV = 8.1 * 10^-19 J

x = 8.1 * 10^-19/ 1.60 x 10-19

x = 5.1 eV

Learn more about cutoff frequency:https://brainly.com/question/14378802

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