Question

The maximum wavelength for photoelectric emission in tungsten is 230 nm. What wavelength of light must be used inorder for electrons with a maximum energy of 1.5 eV to be ejected ?

Answer 1

Answer:

180.04 nm

Explanation:

λ₀ = maximum wavelength for photoelectric emission in tungsten = 230 x 10⁻⁹ m

E₀ = maximum energy of ejected electron = 1.5 eV = 1.5 x 1.6 x 10⁻¹⁹ J

λ = wavelength of light used = ?

Using conservation of energy

Energy of the light used = Maximum energy required for photoelectric emission + Energy of ejected electron

$\frac{hc}{\lambda }=\frac{hc}{\lambda_{o} } + E_{o}$

$\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{\lambda }=\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{230 \times 10^{-9} } + 1.5 \times 1.6 \times 10^{-19}$

λ = 180.04 x 10⁻⁹ m

λ = 180.04 nm