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

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(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \, eV \cdot Å$, we have $E = \frac{12400

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$A$ and $B$

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(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \, eV \cdot Å$, we have $E = \frac{12400}{4000} \, eV = 3.1 \, eV$.Photoelectrons are emitted if the energy of the incident light is greater than or equal to the work function $(\Phi)$ of the material.Comparing $E = 3.1 \, eV$ with the given work functions:For $A$: $1.2 \, eV For $B$: $2.4 \, eV For $C$: $3.6 \, eV > 3.1 \, eV$ (Does not emit)For $D$: $4.8 \, eV > 3.1 \, eV$ (Does not emit)For $E$: $6.0 \, eV > 3.1 \, eV$ (Does not emit)Therefore, only elements $A$ and $B$ will emit photoelectrons.

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 $4000 \, Å$ is allowed to fall on these elements, then photoelectrons are emitted by:

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