KCET · Physics · Dual Nature of Matter
When a piece of metal is illuminated by a monochromatic light of wavelength \(\lambda\), then stopping potential is \(3 \mathrm{~V}_{\mathrm{s}}\). When same surface is illuminated by light of wavelength \(2 \lambda\), then stopping potential becomes \(\mathrm{V}_{\mathrm{s}}\). The value of threshold wavelength for photoelectric emission will be
- A \(4 \lambda\)
- B \(8 \lambda\)
- C \(\frac{4}{3} \lambda\)
- D \(6 \lambda\)
Answer & Solution
Correct Answer
(A) \(4 \lambda\)
Step-by-step Solution
Detailed explanation
According to Einstein's photoelectric equation
\[
\mathrm{eV}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
\]
Ist case
\[
3 \mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
\]
IInd case
\[
\mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right]
\]
Dividing Eq. (i) by Eq. (ii), we get
\[
\lambda_{0}=4 \lambda
\]
\[
\mathrm{eV}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
\]
Ist case
\[
3 \mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
\]
IInd case
\[
\mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right]
\]
Dividing Eq. (i) by Eq. (ii), we get
\[
\lambda_{0}=4 \lambda
\]
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