Question

What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of velocity 6 × 10⁵ ms⁻¹ from it?

(Mass of electron = 9 × 10⁻³¹ kg
Velocity of light = 3 × 10⁸ ms⁻¹
Planck's constant = 6.626 × 10⁻³⁴ Js
Charge of electron = 1.6 × 10⁻¹⁹ JeV⁻¹)

Options

• 0.9 eV

• 3.1 eV

• 2.1 eV

• 4.0 eV

Answer 1

2.1 eV

Explanation:

E = hv = `"hc"/λ`

E = `(6.626 xx 10^-34 xx 3 xx 10^8)/(4000 xx 10^-10)` = 4.97 × 10⁻¹⁹ J

= `(4.97 xx 10^-19  "J")/(1.6 xx 10^-10  "J eV"^-1)`

= 3.1 eV

K.E. = `1/2"mv"^2`

= `1/2 xx 9 xx 10^-31  "kg" xx (6 xx 10^5  "ms"^-1)^2`

= 1.62 × 10⁻¹⁹ J  .....[1 J = kg.m²s⁻²]

= 1 eV

According to the photoelectric effect,

K.E. = hv − hv₀

hv₀ = hv − K.E.

Work function (W₀) = E − K.E. = 3.1 − 1 = 2.1 eV