Subjective Type

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength ?(b) Find the wavelength of light for which the stopping potential is 2.5 V.

Solution

$$\phi =4$$ $$eV=4\times1.6\times10^{-19}J$$

a) Threshold wavelength = $$\lambda $$
$$\phi =hc/\lambda$$

$$\Rightarrow \dfrac{hC}{\phi}=\dfrac{6.63\times10^{-34}\times3\times10^8}{4\times\times1.6\times10^{-19}}$$

= $$\dfrac{6.63\times3}{6.4}$$ $$\times$$ $$\dfrac { { 10 }^{ -27 } }{ { 10 }^{ -9 } } $$ = $$3.1\times10^{-7}m=310nm.$$

b) Stopping potential is $$2.5$$ $$V$$
$$E=\phi+eV$$

$$\Rightarrow hc/\lambda =4\times1.6\times10^{-19}+1.6\times10^{-19}\times2.5$$

$$\Rightarrow \lambda= \dfrac{6.63\times10^{-34}\times3\times10^8}{\lambda \times1.6 \times 10^{-9}}= 4+2.5$$

$$\Rightarrow \dfrac{6.63\times3\times10^{-26}}{1.6\times10^{-19}\times6.5}=1.9125\times10^{-7}=190nm.$$

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