Photons of energy 2.4 text{ eV} and wavelengt | vedclass

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Question

(A) Let $\phi_0$ be the work function of the metal.According to Einstein's photoelectric equation, the maximum kinetic energy $K_{\max}$ is given by:$

Vedclass · Accepted Answer

$2.1$

Vedclass · Answer

(A) Let $\phi_0$ be the work function of the metal.According to Einstein's photoelectric equation, the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi_0$  --- $(i)$Given that the initial energy of the photon is $E = \frac{hc}{\lambda} = 2.4 	ext{ eV}$.When the wavelength is decreased by $50 \%$, the new wavelength is $\lambda' = \frac{\lambda}{2}$.The new energy of the photon is $E' = \frac{hc}{\lambda'} = \frac{hc}{\lambda / 2} = 2 \left( \frac{hc}{\lambda} ight) = 2 	imes 2.4 	ext{ eV} = 4.8 	ext{ eV}$.The new maximum velocity is $v' = 3v$, so the new maximum kinetic energy is $K'_{\max} = \frac{1}{2} m (3v)^2 = 9 \left( \frac{1}{2} m v^2 ight) = 9 K_{\max}$.Using the photoelectric equation for the second case:$9 K_{\max} = E' - \phi_0 = 4.8 - \phi_0$  --- (ii)From equation $(i)$, $K_{\max} = 2.4 - \phi_0$.Substituting this into equation (ii):$9(2.4 - \phi_0) = 4.8 - \phi_0$$21.6 - 9\phi_0 = 4.8 - \phi_0$$8\phi_0 = 21.6 - 4.8 = 16.8$$\phi_0 = \frac{16.8}{8} = 2.1 	ext{ eV}$.

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