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(B) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{400 \ nm} = 3.105 \ eV$.$(a)$ For the photoelectric effect to

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$b, d$

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(B) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{400 \ nm} = 3.105 \ eV$.$(a)$ For the photoelectric effect to occur,the energy of the incident photon must be greater than the work function $(\phi)$. Here,$E = 3.105 \ eV$,which is less than the work functions of $Zn$ $(3.4 \ eV)$,$Fe$ $(4.8 \ eV)$,and $Ni$ $(5.9 \ eV)$. Therefore,no photoelectrons will be emitted from any metal. Statement $(a)$ is incorrect.$(b)$ Since $E $(c)$ According to Einstein's photoelectric equation,$KE_{max} = E - \phi = h
u - \phi$. If the frequency $
u$ is doubled,$KE_{max}' = 2h
u - \phi$. This is not equal to $2(h
u - \phi)$. Statement $(c)$ is incorrect.$(d)$ If $\lambda  \frac{1242}{200} = 6.21 \ eV$. Since $6.21 \ eV$ is greater than the work functions of all three metals $(3.4 \ eV, 4.8 \ eV, 5.9 \ eV)$,photoelectrons will be emitted from all three metals. Statement $(d)$ is correct.Thus,the correct statements are $(b)$ and $(d)$.

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