(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Given $\lambda_{B} = 2 \lambd

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Given $\lambda_{B} = 2 \lambda_{A}$,we have $\frac{h}{\sqrt{2m T_{B}}} = 2 \frac{h}{\sqrt{2m T_{A}}}$.Squaring both sides,we get $\frac{1}{T_{B}} = \frac{4}{T_{A}}$,which implies $T_{A} = 4 T_{B}$.We are given $T_{B} = T_{A} - 1.5$. Substituting $T_{A} = 4 T_{B}$,we get $T_{B} = 4 T_{B} - 1.5$.This simplifies to $3 T_{B} = 1.5$,so $T_{B} = 0.5 \; eV$.Using Einstein's photoelectric equation for metal $B$: $K_{max} = E - \phi_{B}$,where $E = 4.5 \; eV$.$0.5 = 4.5 - \phi_{B}$.Therefore,$\phi_{B} = 4.5 - 0.5 = 4.0 \; eV$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

Medium

When a 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 wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

JEE MAIN 2020 All JEE MAIN

A
$3$
B
$2$
C
$4$
D
$1.5$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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