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A metal exposed to light of wavelength $800 \mathrm{~nm}$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $500 \mathrm{~nm}$ is used. The workfunction of the metal is : (Take $\mathrm{h c=1230 \mathrm{eV}-\mathrm{nm}}$ ).

Option: 1
$1.537 \mathrm{eV}$

Option: 2
$2.46 \mathrm{eV}$

Option: 3
$0.615 \mathrm{eV}$

Option: 4
$1.23 \mathrm{eV}$

Answers (1)

best_answer

$\mathrm{\lambda =800\: nm}$

$\mathrm{E\left ( eV \right )=\frac{hc}{\lambda }}$

$\mathrm{\lambda =\frac{1230}{E}nm}$

$\mathrm{E_{1}=\frac{1230}{\lambda _{1}}=\frac{1230}{800}eV}\rightarrow (1)$

$\mathrm{E_{2}=\frac{1230}{\lambda _{2}}=\frac{1230}{500}eV}\rightarrow (2)$

From Einstein's photoelectric equation,

$\mathrm{E=\phi_0+\left(k E_{\max }\right)}$

Let,

$\mathrm{\left(K E_{\text {max }}\right)_1=K_0}$

$\mathrm{E_1=\phi_0+k_0 \rightarrow \text { (3) }}$

$\mathrm{E_2=\phi_0+2k_0 \rightarrow \text { (4) }(Given)}$

$2\times (3)-(4)$

$\mathrm{2E_{1}-E_{2}=\phi _{0}}$

$\mathrm{\frac{1230}{400}-\frac{1230}{500}=\phi_0}$

$\mathrm{\phi_0=\frac{1230}{2000}=\frac{123}{200} }$

Work Function $\mathrm{=\phi _{0}=0.615eV}$

The work function of thermal is $\mathrm{0.615\: eV}$

Hence (3) is correct option

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