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

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

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

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