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Difficulty: HardWave-Particle Duality and de Broglie Wavelength

A proton of mass 1.67×10−27 kg1.67 \times 10^{-27}\text{ kg} and an electron of mass 9.11×10−31 kg9.11 \times 10^{-31}\text{ kg}, both carrying charges of equal magnitude, are accelerated from rest through the same electric potential difference. Calculate the ratio of the de Broglie wavelength of the electron to that of the proton.

Answer: 42.8

Answer

The ratio of the de Broglie wavelength of the electron to that of the proton is 42.8.
The de Broglie wavelength of a particle accelerated through potential difference VV is given by λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}. Since both particles carry equal charge qq and experience the same potential VV, the ratio of their wavelengths simplifies to λeλp=mpme=1.67×10−279.11×10−31≈42.8\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8.

Step-by-Step Solution

1
Relate kinetic energy to accelerating potential difference
Ek=qVE_k = qV
Electric potential energy converted into kinetic energy during acceleration from rest.
2
Express de Broglie wavelength in terms of particle mass, charge, and potential difference
λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}
Combining p=2mEkp = \sqrt{2mE_k} with de Broglie's formula λ=hp\lambda = \frac{h}{p}.
3
Formulate the wavelength ratio of electron to proton
λeλp=mpme\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}
Planck's constant hh, elementary charge qq, and potential difference VV are identical for both particles and cancel out.
4
Substitute numerical values and compute final ratio
1.67×10−279.11×10−31≈42.8\sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8
Square root of the proton-to-electron mass ratio yields the inverse ratio of their wavelengths.

Key Concept

De Broglie wavelength relation to particle mass under constant accelerating potential
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Topics

Wave-Particle Duality and de Broglie Wavelength

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