Five elements A, B, C, D and E have work func | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400\, eV\cdot\mathring{A}$,we get $E = \frac{12400}{400

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$A$ and $B$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400\, eV\cdot\mathring{A}$,we get $E = \frac{12400}{4000} = 3.1\, eV$.Photoelectric emission occurs only if the energy of the incident photon is greater than the work function of the metal surface $(E > \Phi)$.Comparing the photon energy $(3.1\, eV)$ with the given work functions:For $A$ $(1.2\, eV)$: $3.1 > 1.2$ (Emission occurs)For $B$ $(2.4\, eV)$: $3.1 > 2.4$ (Emission occurs)For $C$ $(3.6\, eV)$: $3.1 For $D$ $(4.8\, eV)$: $3.1 For $E$ $(6.0\, eV)$: $3.1 Therefore,photoelectrons are emitted only by elements $A$ and $B$.

Five elements $A, B, C, D$ and $E$ have work functions $1.2\, eV, 2.4\, eV, 3.6\, eV, 4.8\, eV$ and $6\, eV$ respectively. If light of wavelength $4000\, \mathring{A}$ is allowed to fall on these elements,then photoelectrons are emitted by

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