An electron and a photon each have a wavelength of 1.00 nm. Find
(a)    their momenta,
(b)    the energy of the photon, and
(c)    the kinetic energy of electron.
(Take h = 6.63 x 10^–34 Js).

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Solution

Given,
Wavelength of electron and photon, λ_e = λ_p = 1.00 nm = 1 x 10^–9 m

(a) Momenta of the particle,

p = \frac{h}{λ} = \frac{6.63 \times 10^{- 34}}{1.00 \times 10^{- 19}} kg {ms}^{- 1}

= 6.63 \times 10^{- 25} kg {ms}^{- 1}

(b) Energy of the photon,

E = hv = \frac{hc}{λ}

= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1 \times 10^{- 9}} J

= 1.989 \times 10^{- 16} J = \frac{1.989 \times 10^{- 16}}{1.6 \times 10^{- 19}} eV

= 1.243 ke V

\frac{h}{\sqrt{2 {mE}_{k}}} = λ

\Rightarrow

\sqrt{2 {mE}_{k}} = \frac{h}{λ}

i.e.,

E_{k} = \frac{h^{2}}{2 {mλ}^{2}}

\Rightarrow

E_{k} = \frac{(6.63 \times 10^{- 34})^{2}}{2 \times 9.1 \times 10^{- 31} \times (10^{- 9})^{2}}

= \frac{43.96 \times 10^{- 68}}{18.2 \times 10^{- 49}} J = 2.4 \times 10^{- 19} J

E_{k} = \frac{2.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV = 1.5 eV .

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