Question

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

• A. $A,B$ and $C$
• B. $A,B,C,D$ and $E$
• C. $A$ and $B$
• D. Only $E$

Answer 1

The correct option is C $A$ and $B$

Energy of the photon associated with wavelength $\lambda =4000$ $\stackrel{o}{A}$ is given by
$E_p=\frac{hc}{\lambda }\Rightarrow E_p=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}=3.1\text{eV}$
For photoelectric effect to happen $E_p>\text{work function}\left({\varphi }_0\right)$
So for only A and B will show photoelectric effect as their work function is less than $E_p$.