Question

A photon of wavelength 4 × 10^–7 m strikes on the metal surface, the work function of the metal is 2.13 eV. Calculate
(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10^–19 J).

Answer 1

(i). Energy (E) of a photon = hν= hcλ
Where, h = Planck’s constant = 6.626 × 10–34 Js c = velocity of light in vacuum = 3 × 108 m/s
λ = wavelength of photon = 4 × 10–7 m
Substituting the values in the given expression of E:
E=(6.626×10−34)(3×108)4×10−7=4.9695×10−19J
The energy of the photon is 4.9695×10−19J

Hence, the energy of the photon in eV = 4.9695×10−19J1.6020×10−19J/eV

= 3.1020 eV

(ii). The kinetic energy of emission Ek is given by
hv−hv0=(E−W)eV=(4.9695×10−191.6020×10−19)eV−2.13eV=(3.1020−2.13)eV=0.9710eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii). The velocity of a photoelectron (ν) can be calculated by the expression,
12mv2=hv−hv0⇒v=√2(hv−hv0)m
Where (hv−hv0) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:
v=√2×(0.9720×1.6020×10−19)J9.10939×10−31kg=√0.3418×1012m2s−2
v=5.84×105 ms–1
Hence, the velocity of the photoelectron is 5.84 × 105ms–1.