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

180.04 nm

Step-by-step 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

`(hc)/(\lambda )=(hc)/(\lambda_(o) ) + E_(o)`

`((6.63 * 10^(-34))(3 * 10^(8)))/(\lambda )=((6.63 * 10^(-34))(3 * 10^(8)))/(230 * 10^(-9) ) + 1.5 * 1.6 * 10^(-19)`

λ = 180.04 x 10⁻⁹ m

λ = 180.04 nm