Question

An electron is accelerated from rest through a potential difference. After acceleration the electron has a wavelength of 880 nm. What is the potential difference responsible for the acceleration of the electron? (h = 6.626 × 10-34 J ∙ s, melectron = 9.11 × 10-31 kg, e = 1.6 10-19 C)

1.7 × 10-6 V

1.9 × 10-6 V

2.2 × 10-6 V

2.5 × 10-6 V

Answer 1

As we know that;

= 1/2mv^2=eV

= 1.9*10^-6 V...... Answer

Answer 2

1. First, according to the de Broglie wavelength formula \(\lambda=\frac{h}{p}\), where \(\lambda\) is the wavelength, h is the Planck's constant, and p is the momentum, we can find the momentum p of the electron.

• \(p=\frac{6.626\times10^{-34}}{880\times10^{-9}}=\frac{6.626\times10^{-34}}{8.8\times10^{-7}} = 7.53\times10^{-28}\space kg\cdot m/s\).

• Given that \(\lambda = 880\space nm = 880\times10^{-9}\space m\) and \(h = 6.626\times10^{-34}\space J\cdot s\), from \(p=\frac{h}{\lambda}\), we have:

2. Then, according to the relationship between kinetic energy and momentum \(K=\frac{p^{2}}{2m}\), where K is the kinetic energy and m is the mass, we can find the kinetic energy K of the electron.

• Given that \(m = m_{electron}=9.11\times10^{-31}\space kg\) and \(p = 7.53\times10^{-28}\space kg\cdot m/s\), then \(K=\frac{(7.53\times10^{-28})^{2}}{2\times9.11\times10^{-31}}\)\(= 3.11\times10^{-25}\space J\).

3. Finally, according to the fact that when an electron is accelerated in an electric field, the increase in kinetic energy is equal to the work done by the electric field \(K = qV\), where q is the charge and V is the potential difference, we can find the potential difference V.

• \(V=\frac{3.11\times10^{-25}}{1.6\times10^{-19}}=\frac{3.11}{1.6}\times10^{-6}\)\(\approx 1.9\times10^{-6}\space V\).

• Given that \(q = e = 1.6\times10^{-19}\space C\) and \(K = 3.11\times10^{-25}\space J\), from \(V=\frac{K}{q}\), we have:

So the potential difference that causes the electron to accelerate is \(1.9\times 10^{-6}\space V\). The answer is \(1.9\times 10^{-6}\space V\)