Physics · 2020
JEE Main · 8 January 2020, Shift 1 · Q17
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A eV and de-Broglie…
When photon of energy $\displaystyle 4.0$ eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $\displaystyle \mathrm{T}_{\mathrm{A}} \mathrm{eV}$ and de-Broglie wavelength $\displaystyle \lambda_{\mathrm{A}}$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy $\displaystyle 4.50$ eV is $\displaystyle \mathrm{T}_{\mathrm{B}}=\left(\mathrm{T}_{\mathrm{A}}-1.5\right) \mathrm{eV}$. If the de-Broglie wavelength of these photoelectrons $\displaystyle \lambda_{\mathrm{B}}=2 \lambda_{\mathrm{A}}$, then the work function of metal $\displaystyle B$ is :
Official answer
(2)
$\displaystyle 4$ eV
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.