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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 \(=6.63 \times 10^{-34} \; Js\) )
- A \(32\)
- B \(33\)
- C \(34\)
- D \(35\)
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 \(\phi= E - eV _{0}\) \(E =\frac{12400}{6630} eV =1.87 \; eV\) \(\phi=(1.87-0.42)=1.45 \; eV\) \(\phi= hv _{0} ; v _{0}\)…
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