1. λ₀/(1+eE₀/mV₀t) 2. λ₀ t 3. λ₀(1+eE₀/mV₀t) 4. λ₀

Answer: 1. λ₀/(1+eE₀/mV₀t) The electron has charge -e and the field is -E₀hat i , so the force is +eE₀hat i . It accelerates along its motion with a=eE₀/m . v=V₀+eE₀/mt=V₀(1+eE₀t/mV₀) de-Broglie wavelength λ=h/p∝1/v , so λ=λ₀/1+eE₀/mV₀t

Topic: Matter waves, de Broglie relation· Dual Nature of Matter and Radiation

An electron of mass m with an initial velocity V=V_0 i (V_0>0) enters an electric field E=-E_0 i (E_0 = constant >0) at t = 0. If λ_0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

1. $\dfrac{\lambda_0}{\left(1+\frac{eE_0}{mV_0}t\right)}$

2. $\lambda_0 t$

3. $\lambda_0\left(1+\frac{eE_0}{mV_0}t\right)$

4. $\lambda_0$

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Answer: 1. $\dfrac{\lambda_0}{\left(1+\frac{eE_0}{mV_0}t\right)}$

The electron has charge $-e$ and the field is $-E_0\hat i$, so the force is $+eE_0\hat i$. It accelerates along its motion with $a=\dfrac{eE_0}{m}$.

$$v=V_0+\frac{eE_0}{m}t=V_0\left(1+\frac{eE_0t}{mV_0}\right)$$

de-Broglie wavelength $\lambda=\dfrac hp\propto\dfrac1v$, so

$$\lambda=\frac{\lambda_0}{1+\dfrac{eE_0}{mV_0}t}$$

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An electron of mass m with an initial velocity V=V_0 i (V_0>0) enters an electric field E=-E_0 i (E_0 = constant >0) at t = 0. If λ_0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

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