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Home » An electron and a photon each have a wavelength of nm. Find: (a) Their momenta, (b) The energy of the photon
An electron and a photon each have a wavelength of 1.00 nm. Find: (a) Their momenta, (b) The energy of the photon
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
An electron and a photon each have a wavelength of 1.00 nm. Find: (a) Their momenta, (b) The energy of the photon

Wavelength of an electron is represented by \lambda_{e} and that of a photon by \lambda_{p},

So,

\lambda_{e}=\lambda_{p}=\lambda=1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}

Planck’s constant, h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}

(a) The momentum of an elementary particle is given by the following de Broglie relation:

\lambda=\frac{h}{p}

p=\frac{h}{\lambda}

It is obvious that momentum is solely determined by the particle’s wavelength. Both an electron and a photon have the same momentum since their wavelengths are the same.

Therefore, p=\frac{6.63 \times 10^{-34}}{1 \times 10^{-9}}

=6.63 \times 10^{-25} \mathrm{Kg} \mathrm{m} / \mathrm{s}

(b) The energy of a photon is given by the relation:

\mathrm{E}=\frac{h c}{\lambda}

Where,

Speed of light, c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}

Therefore, E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1 \times 10^{-9} \times 1.6 \times 10^{-1.3}}

=1243.1 \mathrm{eV}=1.243 \mathrm{keV}

Therefore, the energy of the photon is 1.243 \mathrm{keV} .

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