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(D) Energy of incident photon $E = \frac{1240 \ eV \cdot nm}{\lambda (nm)} = \frac{1240}{124} = 10 \ eV$. Work function $\Phi = 4 \ eV$. Maximum kinet

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$0.5$

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(D) Energy of incident photon $E = \frac{1240 \ eV \cdot nm}{\lambda (nm)} = \frac{1240}{124} = 10 \ eV$. Work function $\Phi = 4 \ eV$. Maximum kinetic energy $K.E._{max} = E - \Phi = 10 \ eV - 4 \ eV = 6 \ eV$. The de Broglie wavelength $\lambda$ of an electron is given by $\lambda = \frac{h}{\sqrt{2mK.E.}} \approx \sqrt{\frac{150}{K.E. (eV)}} \ \mathring{A}$. $\lambda = \sqrt{\frac{150}{6}} \ \mathring{A} = \sqrt{25} \ \mathring{A} = 5 \ \mathring{A}$. Since $1 \ nm = 10 \ \mathring{A}$,$\lambda = 0.5 \ nm$.

The wavelength of the electron emitted by a metal sheet of work function $4 \ eV$ when photons from $EMR$ of wavelength $124 \ nm$ strike the metal plate is ........... $nm$.

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