JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2026
- Medium
- MCQ
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 .
\(\left( h =6.63 \times 10^{-34} J . s , e =1.6 \times 10^{-19} C , c =3 \times 10^8 m / s \right)\)
Pick an option to check your answer.
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)
\(q.(2.5)=\frac{hc}{2\lambda}\)
\(2.5=(\frac{hc}{e})(\frac{1}{2\lambda})\)
\(2.5=\frac{12400}{2(\lambda)}\)
\(\lambda=\frac{12400}{5}\text{\AA}\)
\(\lambda=2480~\text{\AA}\)
\(\lambda=2.48\times10^{-7}m\)
\(q(0.7)=\frac{hc}{2\lambda}-\phi\) ...(2)
Eq. (1) - Eq. (2)
\(q.(2.5)=\frac{hc}{2\lambda}\)
\(2.5=(\frac{hc}{e})(\frac{1}{2\lambda})\)
\(2.5=\frac{12400}{2(\lambda)}\)
\(\lambda=\frac{12400}{5}\text{\AA}\)
\(\lambda=2480~\text{\AA}\)
\(\lambda=2.48\times10^{-7}m\)
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