Answer to Question #171800 in General Chemistry for ncssr

Question

The threshold wavelength for photoelectric emission in Tungsten is 240 nm. What wavelength of light must be used in order that the emitted photoelectrons have a maximum K.E. of 1.2 eV (h = 4.14*10^-15 eV-s, c = 3*10^8 m/s)

Answer 1

work function = $\dfrac{hc}{\lambda_o} = \dfrac{4.14×10^-¹⁵×3×10⁸}{240 × 10⁹} = 5.175\ eV$

E = K.E_max + W

E = 1.2 + 5.175 = 6.375 eV

$\dfrac{hc}{\lambda_{max}} = 6.375$

$\dfrac{1.242×10^-⁶}{\lambda_{max}} = 6.375$

$\lambda_{max} = \dfrac{1.242×10^{-6}}{6.375} = 1.89 × 10^{-7} = 189nm$