The work function of a certain metal is $3.31 \times 10^{-19} \mathrm{~J}$. Then, the maximum kinetic energy…

The work function of a certain metal is $3.31 \times 10^{-19} \mathrm{~J}$. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 Ã…$ is $\begin{aligned} & \text { (Given, } h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}, c=3 \times 10^8 \mathrm{~ms}^{-1} \text {, } \\ & e=1.6 \times 10^{-19} \mathrm{C} \text { ) } \end{aligned}$
  1. $2.48 \mathrm{eV}$
  2. $0.41 \mathrm{eV}$
  3. $2.07 \mathrm{eV}$
  4. $0.82 \mathrm{eV}$

Solution

Work function $W_0=3.31 \times 10^{-19} \mathrm{~J}$ Wavelength of incident radiation $\begin{aligned} & \lambda=5000 \times 10^{-10} \mathrm{~m} \\ & E=W_0+\mathrm{KE} \end{aligned}$ (According to Einstein equation) $\begin{aligned} \frac{h c}{\lambda} & =3.31 \times 10^{-19}+\mathrm{KE} \\ \mathrm{KE} & =-3.31 \times 10^{-19}+\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} \\ & =-3.31 \times 10^{-19}+\frac{6.62 \times 3}{5} \times 10^{-19} \\ & =(-3.31 \times 1.324 \times 3) \times 10^{-19} \\ & =(3.972-3.31) \times 10^{-19} \\ & =0.662 \times 10^{-19} \mathrm{~J} \end{aligned}$ $\begin{aligned} \Rightarrow E & =\frac{0.662 \times 10^{-19}}{1.6 \times 10^{-19}} \\ & =0.41 \mathrm{eV}\end{aligned}$

Asked in: AP EAMCET 2009

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