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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength \(491\, nm\) is \(0.710\, V\). When the incident wavelength is changed to a new value, the stopping potential is \(1.43\, V\). The new wavelength is ....... \(nm.\)
- A \(329\)
- B \(309\)
- C \(382\)
- D \(400\)
Answer & Solution
Correct Answer
(C) \(382\)
Step-by-step Solution
Detailed explanation
\(\frac{ hc }{\lambda}=\phi+ eV _{ s }\) \(\frac{1240}{491}=\phi+0.71\) \(\frac{1240}{\lambda}=\phi+1.43 \) \(\therefore \lambda=382\, nm\)
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