La longitud de onda de De Broglie asociada a una partícula es:
λ = h p = h m v \lambda = \frac{h}{p} = \frac{h}{mv} λ = p h = m v h Una partícula de carga e e e acelerada mediante una diferencia de potencial V V V adquiere energía cinética:
e V = 1 2 m v 2 ⟹ m v = 2 m e V eV = \frac{1}{2}mv^2 \implies mv = \sqrt{2meV} e V = 2 1 m v 2 ⟹ m v = 2 m e V Por tanto, la longitud de onda de De Broglie queda:
λ = h 2 m e V \lambda = \frac{h}{\sqrt{2meV}} λ = 2 m e V h Para el electrón ( m e = 9,11 ⋅ 10 − 31 m_e = 9{,}11 \cdot 10^{-31} m e = 9 , 11 ⋅ 1 0 − 31 kg):
λ e = h 2 m e e V = 6,63 ⋅ 10 − 34 2 ⋅ 9,11 ⋅ 10 − 31 ⋅ 1,60 ⋅ 10 − 19 ⋅ 2 ⋅ 10 4 \lambda_e = \frac{h}{\sqrt{2\, m_e\, e\, V}} = \frac{6{,}63 \cdot 10^{-34}}{\sqrt{2 \cdot 9{,}11 \cdot 10^{-31} \cdot 1{,}60 \cdot 10^{-19} \cdot 2 \cdot 10^{4}}} λ e = 2 m e e V h = 2 ⋅ 9 , 11 ⋅ 1 0 − 31 ⋅ 1 , 60 ⋅ 1 0 − 19 ⋅ 2 ⋅ 1 0 4 6 , 63 ⋅ 1 0 − 34 Calculamos el radicando del denominador para el electrón:
2 ⋅ 9,11 ⋅ 10 − 31 ⋅ 1,60 ⋅ 10 − 19 ⋅ 2 ⋅ 10 4 = 5,831 ⋅ 10 − 45 kg 2 ⋅ m 2 ⋅ s − 2 2 \cdot 9{,}11 \cdot 10^{-31} \cdot 1{,}60 \cdot 10^{-19} \cdot 2 \cdot 10^{4} = 5{,}831 \cdot 10^{-45} \text{ kg}^2 \cdot \text{m}^2 \cdot \text{s}^{-2} 2 ⋅ 9 , 11 ⋅ 1 0 − 31 ⋅ 1 , 60 ⋅ 1 0 − 19 ⋅ 2 ⋅ 1 0 4 = 5 , 831 ⋅ 1 0 − 45 kg 2 ⋅ m 2 ⋅ s − 2 5,831 ⋅ 10 − 45 = 7,636 ⋅ 10 − 23 kg ⋅ m ⋅ s − 1 \sqrt{5{,}831 \cdot 10^{-45}} = 7{,}636 \cdot 10^{-23} \text{ kg} \cdot \text{m} \cdot \text{s}^{-1} 5 , 831 ⋅ 1 0 − 45 = 7 , 636 ⋅ 1 0 − 23 kg ⋅ m ⋅ s − 1 λ e = 6,63 ⋅ 10 − 34 7,636 ⋅ 10 − 23 ≈ 8,68 ⋅ 10 − 12 m \lambda_e = \frac{6{,}63 \cdot 10^{-34}}{7{,}636 \cdot 10^{-23}} \approx 8{,}68 \cdot 10^{-12} \text{ m} λ e = 7 , 636 ⋅ 1 0 − 23 6 , 63 ⋅ 1 0 − 34 ≈ 8 , 68 ⋅ 1 0 − 12 m Para el protón ( m p = 1,67 ⋅ 10 − 27 m_p = 1{,}67 \cdot 10^{-27} m p = 1 , 67 ⋅ 1 0 − 27 kg):
λ p = h 2 m p e V = 6,63 ⋅ 10 − 34 2 ⋅ 1,67 ⋅ 10 − 27 ⋅ 1,60 ⋅ 10 − 19 ⋅ 2 ⋅ 10 4 \lambda_p = \frac{h}{\sqrt{2\, m_p\, e\, V}} = \frac{6{,}63 \cdot 10^{-34}}{\sqrt{2 \cdot 1{,}67 \cdot 10^{-27} \cdot 1{,}60 \cdot 10^{-19} \cdot 2 \cdot 10^{4}}} λ p = 2 m p e V h = 2 ⋅ 1 , 67 ⋅ 1 0 − 27 ⋅ 1 , 60 ⋅ 1 0 − 19 ⋅ 2 ⋅ 1 0 4 6 , 63 ⋅ 1 0 − 34 Calculamos el radicando del denominador para el protón:
2 ⋅ 1,67 ⋅ 10 − 27 ⋅ 1,60 ⋅ 10 − 19 ⋅ 2 ⋅ 10 4 = 1,069 ⋅ 10 − 41 kg 2 ⋅ m 2 ⋅ s − 2 2 \cdot 1{,}67 \cdot 10^{-27} \cdot 1{,}60 \cdot 10^{-19} \cdot 2 \cdot 10^{4} = 1{,}069 \cdot 10^{-41} \text{ kg}^2 \cdot \text{m}^2 \cdot \text{s}^{-2} 2 ⋅ 1 , 67 ⋅ 1 0 − 27 ⋅ 1 , 60 ⋅ 1 0 − 19 ⋅ 2 ⋅ 1 0 4 = 1 , 069 ⋅ 1 0 − 41 kg 2 ⋅ m 2 ⋅ s − 2 1,069 ⋅ 10 − 41 = 1,034 ⋅ 10 − 20,5 ≈ 1,034 ⋅ 10 − 20 ⋅ 10 − 0,5 ≈ 3,270 ⋅ 10 − 21 kg ⋅ m ⋅ s − 1 \sqrt{1{,}069 \cdot 10^{-41}} = 1{,}034 \cdot 10^{-20{,}5} \approx 1{,}034 \cdot 10^{-20} \cdot 10^{-0{,}5} \approx 3{,}270 \cdot 10^{-21} \text{ kg} \cdot \text{m} \cdot \text{s}^{-1} 1 , 069 ⋅ 1 0 − 41 = 1 , 034 ⋅ 1 0 − 20 , 5 ≈ 1 , 034 ⋅ 1 0 − 20 ⋅ 1 0 − 0 , 5 ≈ 3 , 270 ⋅ 1 0 − 21 kg ⋅ m ⋅ s − 1 λ p = 6,63 ⋅ 10 − 34 3,270 ⋅ 10 − 21 ≈ 2,03 ⋅ 10 − 13 m \lambda_p = \frac{6{,}63 \cdot 10^{-34}}{3{,}270 \cdot 10^{-21}} \approx 2{,}03 \cdot 10^{-13} \text{ m} λ p = 3 , 270 ⋅ 1 0 − 21 6 , 63 ⋅ 1 0 − 34 ≈ 2 , 03 ⋅ 1 0 − 13 m La relación entre ambas longitudes de onda se obtiene directamente a partir de la expresión λ = h 2 m e V \lambda = \dfrac{h}{\sqrt{2meV}} λ = 2 m e V h , dado que h h h , e e e y V V V son comunes a ambas:
λ e λ p = 2 m p e V 2 m e e V = m p m e = 1,67 ⋅ 10 − 27 9,11 ⋅ 10 − 31 = 1834 ≈ 42,8 \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2\, m_p\, e\, V}}{\sqrt{2\, m_e\, e\, V}} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1{,}67 \cdot 10^{-27}}{9{,}11 \cdot 10^{-31}}} = \sqrt{1834} \approx 42{,}8 λ p λ e = 2 m e e V 2 m p e V = m e m p = 9 , 11 ⋅ 1 0 − 31 1 , 67 ⋅ 1 0 − 27 = 1834 ≈ 42 , 8 Por tanto, la longitud de onda del electrón es aproximadamente 42,8 veces mayor que la del protón cuando ambos son acelerados con la misma diferencia de potencial. Esto se debe a que el protón, al ser mucho más masivo, adquiere un momento lineal mayor y, en consecuencia, una longitud de onda de De Broglie menor.
