The work function of caesium is 2.14, eV. Fin | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = e V_0$,where $V_0$ is t

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None of these

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = e V_0$,where $V_0$ is the stopping potential.Einstein's equation is $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Given: $\phi = 2.14\, eV$,$V_0 = 0.60\, V$,so $K_{max} = 0.60\, eV$.Substituting the values: $0.60\, eV = \frac{1240\, eV \cdot nm}{\lambda} - 2.14\, eV$.$0.60 + 2.14 = \frac{1240}{\lambda}$.$2.74 = \frac{1240}{\lambda}$.$\lambda = \frac{1240}{2.74} \approx 452.55\, nm$.Since $452.55\, nm$ is not among the options,the correct choice is $D$.

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 $0.60\, V$. (Result in $nm$)

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