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 $4.97 × 10^{19} J$.

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(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 $0.97\ eV$.

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(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 $5.84×10^5\ ms^{-1}.$