The work function of caesium is 2.14 eV. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of 0.60 volt :- (1) 454 nm (2) 640 nm (3) 540 nm (4) None of these

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Please find below the solution to the asked query:

Given that the stopping potential for the cesium is 0.6 V, work function is 2.14 eV. So, from the equation of Einstein's photoelectric equation,

$$\begin{gathered} e{V}_s=\frac{hc}{\lambda }-\varphi \Rightarrow 0.6 eV=\frac{12400}{\lambda \left(in {\displaystyle \stackrel{0}{A}}\right)}-2.14 \Rightarrow \frac{12400}{\lambda \left(in {\displaystyle \stackrel{0}{A}}\right)}=2.14+0.6 \\ \Rightarrow \frac{12400}{\lambda \left(in {\displaystyle \stackrel{0}{A}}\right)}=2.74 \Rightarrow \lambda \left(in \stackrel{0}{A}\right)=\frac{12400}{2.74} \\ \Rightarrow \lambda =4525.5 \stackrel{0}{A} \end{gathered}$$

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