ExamBro
ExamBro
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.
Verified Solution

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

\(1.45 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} \times v _{0}\)

\(v _{0}=0.35 \times 10^{15}\)

\(=35 \times 10^{13} sec ^{-1}\)

\(=35\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app