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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2022
- Medium
- Numerical
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength \(6630 \; \mathring A\) is \(0.42 \; V\). If the threshold frequency is \(x \times\) \(10^{13} / s\), where \(x\) is\(\dots\) (nearest integer). (Given, speed light \(=3 \times 10^{8} \; m / s\), Planck's constant \(=6.63 \times 10^{-34} \; Js\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(35\)
Step-by-step Solution
Detailed explanation
Stopping potential \(V _{0}=0.42 \; V\)
\(\lambda=6630 \; \mathring A\)
\(E =\phi+ eV _{0}\)
\(E\) : energy of incident photon
\(V_{0}:\) Stopping potential
\(\phi= E - eV _{0}\)
\(E =\frac{12400}{6630} eV =1.87 \; eV\)
\(\phi=(1.87-0.42)=1.45 \; eV\)
\(\phi= hv _{0} ; v _{0}\) : threshold frequency
\(1.45 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} \times v _{0}\)
\(v _{0}=0.35 \times 10^{15}\)
\(=35 \times 10^{13} sec ^{-1}\)
\(=35\)
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