Dual Nature of Matter

JEE Main level, single correct, medium

172 of 200

Question 172 of 200

In a photoelectric effect experiment, the threshold wavelength of light is380nm380nm . If the wavelength of incident light is260nm260nm , the maximum kinetic energy of emitted electrons will be
Given EE (in eVeV ) =1237λ(innm)=\dfrac{1237}{\lambda (innm)}

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Answer: (C) 1.5eV1.5eV

From Einstein’s photoelectric equation,
K.Emax=hcλ−hcλ0K.{E}_{max}=\dfrac{hc}{\lambda }-\dfrac{hc}{{\lambda }_{0}} (here, λ0={\lambda }_{0}= threshold wavelength)
=1237260−1237380=\dfrac{1237}{260}-\dfrac{1237}{380}
=1237(380−260380×260)=1237\left(\dfrac{380-260}{380\times 260}\right)
=1.5eV=1.5 eV

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