Question
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is $\_\_\_\_$ m.
( $h = 6.63 \times 10^{- 34}\text{ }J.s,e = 1.6 \times 10^{- 19}C,c = 3 \times 10^{8}\text{ }m/s$ )
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Solution
$Q.(3.2) = \frac{hc}{\lambda} - \phi$
$$\begin{array}{r} q(0.7) = \frac{hc}{2\lambda} - \phi\#(1) \end{array}$$
Eq. (1) – Eq. (2)
$${q \cdot (2.5) = \frac{hc}{2\lambda} }{2.5 = \left( \frac{hc}{e} \right)\left( \frac{1}{2\lambda} \right)}$$
$$\begin{matrix} & 2.5 = \frac{12400}{2(\lambda)} \\ & \lambda = \frac{12400}{5}\text{Å} \\ & \lambda = 2480\text{Å} \\ & \lambda = 2.48 \times 10^{- 7}\text{ }m \end{matrix}$$
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