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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2020
- Medium
- Numerical
When photon of energy \(4.0\; eV\) strikes the surface of a metal \(\mathrm{A}\), the ejected photoelectrons have maximum kinetic energy \(\mathrm{T}_{\mathrm{A}}\; eV\) end de-Broglie wavelength \(\lambda_{\mathrm{A}} .\) The maximum kinetic energy of photoelectrons liberated from another metal \(\mathrm{B}\) by photon of energy \(4.50 \;eV\) is \(\mathrm{T}_{\mathrm{B}}=\left(\mathrm{T}_{\mathrm{A}}-1.5\right)\;eV\). If the de-Broglie wavelength of these photoelectrons \(\lambda_{B}=2 \lambda_{A}\) then the work function of metal \(\mathrm{B}\) is ............. \(eV\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\lambda_{\mathrm{B}}=2 \lambda_{\mathrm{A}}\)
\(\Rightarrow \frac{\mathrm{h}}{\sqrt{2 \mathrm{T}_{\mathrm{B}} \mathrm{m}}}=\frac{2 \mathrm{h}}{\sqrt{2 \mathrm{T}_{\mathrm{A}} \mathrm{m}}}\)
\(\mathrm{T}_{\mathrm{A}}=4 \mathrm{T}_{\mathrm{B}}\) and \(\mathrm{T}_{\mathrm{B}}=\left(\mathrm{T}_{\mathrm{A}}-1.5\right)\) eV
\(3 \mathrm{T}_{\mathrm{B}} 1.5 \mathrm{eV} \Rightarrow \mathrm{T}_{\mathrm{B}}=0.5 \mathrm{eV}\)
\(\mathrm{T}_{\mathrm{B}}=0.5 \mathrm{eV}=4.5 \mathrm{eV}-\phi_{\mathrm{B}}\)
\(\phi=4 \mathrm{eV}\)
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