When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is : (Given hc = 1240 eV nm)
- 4 V
- 4.1 V
- 2 V
- 1.5 V
The Correct Option is C
Solution and Explanation
The energy of the incident photon is given by:
\[\frac{hc}{\lambda} = eV_s + \phi\]
Substitute the given values:
\[\frac{1240}{300} \, \text{eV} - 2.13 \, \text{eV} = eV_s\]
\[4.13 \, \text{eV} - 2.13 \, \text{eV} = eV_s\]
\[V_s = 2 \, \text{V}\]
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