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de Broglie wavelength associated with an electron

de Broglie wavelength associated with an electron :- Let an electron is accelerated from rest through a potential difference of V volt, then

Kinetic energy acquired by the electron, $\displaystyle =\frac{1}{2}m{{v}^{2}}$

Work done on electron, $\displaystyle =eV$

$\displaystyle \frac{1}{2}m{{v}^{2}}=eV$

$\displaystyle \Rightarrow v=\sqrt{\frac{2eV}{m}}$

If the de Broglie wavelength of the electron is λ , then…

$\displaystyle \lambda =\frac{h}{mv}=\frac{h}{m\sqrt{2eV/m}}=\frac{h}{\sqrt{2meV}}$

$\displaystyle \Rightarrow \lambda =\frac{6.63\times {{10}^{-34}}}{\sqrt{2\times 9\times {{10}^{-31}}\times 1.6\times {{10}^{-19}}\times V}}=\frac{12.27}{\sqrt{V}}\times {{10}^{-10}}m$

$\displaystyle \lambda =\frac{12.27}{\sqrt{V}}{{A}^{\circ }}$

de Broglie wavelength associated with an electron and other charge particles

The energy of a charge particle accelerated through potential difference V is given by,

$\displaystyle E=\frac{1}{2}m{{v}^{2}}=qV$

Hence de Broglie wavelength

$\displaystyle \lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}$

Using above formula,

For electron (m = 9.1 × 10^-31kg, q = 1.6 × 10^-19 C)

$\displaystyle {{\lambda }_{electron}}=\frac{12.27}{\sqrt{V}}{\AA}$

For Proton (m = 1.67 × 10^-27kg, q = 1.6 × 10^-19 C)

$\displaystyle {{\lambda }_{proton}}=\frac{0.286}{\sqrt{V}}{\AA}$

For Deutron(₁H²) (m = 2 × m_p= 2 × 1.67 × 10^-27kg, q = 1.6 × 10^-19 C)

$\displaystyle {{\lambda }_{deutron}}=\frac{0.202}{\sqrt{V}}{\AA}$

For alpha particle (m = 4 × m_p = 4 × 1.67 × 10^-27kg, q = 2 × 1.6 × 10^-19 C)

$\displaystyle {{\lambda }_{\alpha -particle}}=\frac{0.101}{\sqrt{V}}{\AA}$

Example 1 :- Relation between wavelength of photon and electron of same energy is

(A) $\displaystyle {{\lambda }_{ph}}>{{\lambda }_{e}}$

(B) $\displaystyle {{\lambda }_{ph}}<{{\lambda }_{e}}$

(D) $\displaystyle \frac{{{\lambda }_{e}}}{{{\lambda }_{ph}}}=\text{constant}$

Solution :-

As de Broglie wavelength associated with an electron is given by

$\displaystyle {{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}E}}$ …..(1)

And for photon

$\displaystyle E=h\nu =\frac{hc}{{{\lambda }_{ph}}}$ …..(2)

Using this value in equation (1) we get,

$\displaystyle {{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}\frac{hc}{{{\lambda }_{ph}}}}}=\sqrt{\frac{h{{\lambda }_{ph}}}{2{{m}_{e}}c}}$

$\displaystyle \Rightarrow {{\lambda }_{ph}}=\frac{2{{m}_{e}}c}{h}{{\left( {{\lambda }_{e}} \right)}^{2}}$

Now order of $\displaystyle \frac{2{{m}_{e}}c}{h}$ is about 10¹¹, hence $\displaystyle {{\lambda }_{ph}}>{{\lambda }_{e}}$ .

So option (A) is correct.

Example 2 :- A proton and photon both have same energy of E = 100 keV. The de-Broglie wavelength of proton and photon be λ₁ and λ₂ then λ₁/λ₂ is proportional to

(A) E^-1/2

(B) E^1/2

(D) E

Solution:-

de Broglie wavelength associated with a proton is given by

$\displaystyle {{\lambda }_{1}}=\frac{h}{\sqrt{2{{m}_{p}}E}}$ …..(1)