Question

light of wavelength 6200 Amstrong falls on a metallic having photo electric work function 2 EV what is the value of stopping potential​

Answer 1

Answer:

0 volt

Explanation:

Given:

Wavelength of light=6200Armstrong=620nm

Work function = 2eV

Stopping Potential (Vo) = ?

hc = 1240eV-nm. e = 1.6* 10^-19 J = 1eV

SHORTCUT METHOD:

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