Question

A photon of wavelength \(4×10^{–7} m\) strikes on metal surface, the work function of the metal being \(2.13\ eV\). Calculate:

1. the energy of the photon (eV)
2. the kinetic energy of the emission, and
3. the velocity of the photo electron (\(1\ eV= 1.6020×10^{–19} J\)).

Answer 1

(i) Energy (E) of a photon = \(hν\) = \(\frac {hc}{λ}\)
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:
\(E = \frac {(6.626×10^{-34})(3×10^8)}{4×10^{-7}}\)
\(E= 4.9695×10^{-19}\ J\)

Hence, the energy of the photon is \(4.97 × 10^{19} J\).

---

(ii) The kinetic energy of emission \(K_E\) is given by
\(K_E= hv - hv_0\)
\(K_E= (E - W)\ eV\)
\(K_E= (3.1020-2.13)\ eV\)
\(K_E= 0.9720 \ eV\)
Hence, the kinetic energy of emission is \(0.97\ eV\).

---

(iii) The velocity of a photoelectron (V) can be calculated by the expression,
\(\frac 12 mv^2 = hv - hv_0\)
⇒ \(v = \sqrt {\frac {2(hv - hv_0)}{m}}\)
Where,
\((hv - hv_0)\) 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:
\(v = \sqrt {\frac {2×(0.9720×1.6020×10^{-19})J}{9.10939×10^{-31} kg}}\)
\(v= \sqrt {0.3418×10^{12}\ m^2 s^{-2}}\)
\(v = 5.84×10^5\  ms^{-1}\)

Hence, the velocity of the photoelectron is \(5.84×10^5\  ms^{-1}.\)