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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2020
  • Medium
  • Numerical

The surface of a metal is illuminated altemately with photons of energies \(E _{1}=4\, eV\) and \(E _{2}=2.5\, eV\) respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is \(2 .\) The work function of the metal in \(( eV )\) is\(......ev\)

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Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(E _{1}=\phi+ K _{1} \ldots(1)\)

\(E _{2}=\phi+ K _{2}\) \(...(2)\)

\(E _{1}- E _{2}= K _{1}- K _{2}\)

Now \(\frac{ V _{1}}{ V _{2}}=2\)

\(\frac{ K _{1}}{ K _{2}}=4\)

\(K _{1}=4 K _{2}\)

Now from equation \((2)\)

\(\Rightarrow 4-2.5=4 K _{2}- K _{2}\)

\(1.5=3 K _{2}\)

\(K _{2}=0.5 eV\)

Now putting This

Value in equation \(( 2 )\)

\(2.5=\phi+0.5 eV\)

\(\phi=2 ev\)

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