Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons are

5.5 V

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1.5 V

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9.5 V

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4.0 V

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Solution

The correct option is D
4.0 V

$K_{m a x} = e V_{0} \Rightarrow e V_{0} = 4 e V \Rightarrow V_{0} = 4 V$

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Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons are

The correct option is D 4.0 V Kmax=eV0⇒eV0=4eV⇒V0=4V

The correct option is D
4.0 V

$K_{m a x} = e V_{0} \Rightarrow e V_{0} = 4 e V \Rightarrow V_{0} = 4 V$