JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is _________ m .
\(\left( h =6.63 \times 10^{-34} J . s , e =1.6 \times 10^{-19} C , c =3 \times 10^8 m / s \right)\)
- A \(2.9 \times 10^{-8}\)
- B \(2.2 \times 10^{-8}\)
- C \(3.1 \times 10^{-7}\)
- D \(2.5 \times 10^{-7}\)
Answer & Solution
Correct Answer
(D) \(2.5 \times 10^{-7}\)
Step-by-step Solution
Detailed explanation
\(q.(3.2)=\frac{hc}{\lambda}-\phi\) ...(1) \(q(0.7)=\frac{hc}{2\lambda}-\phi\) ...(2) Eq. (1) - Eq. (2) \(q.(2.5)=\frac{hc}{2\lambda}\) \(2.5=(\frac{hc}{e})(\frac{1}{2\lambda})\) \(2.5=\frac{12400}{2(\lambda)}\) \(\lambda=\frac{12400}{5}\text{\AA}\) \(\lambda=2480~\text{\AA}\)…
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