Question

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what potential difference must an electron [m_0 = 9.11E-31 kg), be accelerated to have a wavelength lambda = 4.23E-12 m?

Answer 1

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.