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  • The de Broglie wavelength of an electron moving with a velocity <img alt="1.5 \times 10^4 \mathrm{~m/s}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.5%20%5Ctimes%2010%5E4%20%5Cmathrm%7B%

The de Broglie wavelength of an electron moving with a velocity 1.5 \times 10^4 \mathrm{~m/s} is equal to that of a photon. The
ratio of the kinetic energy of the electro to the energy of the photon is:

Option: 1

1/4


Option: 2

1/2


Option: 3

2


Option: 4

4


Answers (1)

best_answer

de Broglie wavelength of an electron, \lambda_e=\frac{h}{m_e v_e}

where \mathrm{h} is the Plank's constant, \mathrm{m}_{\mathrm{e}} and \mathrm{v}_{\mathrm{e}}  is the mass and velocity of the electron respectively.
Wavelength of a photon = \lambda_{\mathrm{ph}}

As per question
\lambda_{\mathrm{ph}}=\lambda_a=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}} \quad \quad \quad \quad(i)

Kinetic energy of the electron,
\mathrm{K}_{\mathrm{e}}=\frac{1}{2} \mathrm{~m}_{o} \mathrm{v}_{\mathrm{o}}^2 \quad \quad \quad \quad(ii)
Energy of the photon,
\mathrm{E}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}\quad \quad \quad \quad(iii)
Divide (ii) and (iii), we get

\begin{aligned} & \frac{\mathrm{K}_{e}}{\mathrm{E}_{\mathrm{ph}}}=\frac{\frac{1}{2} \mathrm{~m}_{o} \mathrm{v}_{o}^2}{\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}}=\frac{1}{2} \frac{\mathrm{m}_{o} \mathrm{v}_{\mathrm{o}}^2 \lambda_{\mathrm{ph}}}{\mathrm{hc}}=\frac{1}{2} \frac{\mathrm{m}_{e} \mathrm{v}_{e}^2 \mathrm{~h}}{\mathrm{hcm}_{\mathrm{e}} \mathrm{v}_{e}} \quad \text { (Using (i)) } \\ & =\frac{1}{2} \frac{\mathrm{v}_{e}}{\mathrm{c}}=\frac{1.5 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 3 \times 10^8 \mathrm{~m} / \mathrm{s}}=\frac{1}{4} \end{aligned}

Posted by

Nehul

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