Respuesta :
Explanation:
1. Mass of the proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Wavelength, [tex]\lambda_p=4.23\times 10^{-12}\ m[/tex]
We need to find the potential difference. The relationship between potential difference and wavelength is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}[/tex]
[tex]V=\dfrac{h^2}{2q_pm_p\lambda^2}[/tex]
[tex]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}[/tex]
V = 45.83 volts
2. Mass of the electron, [tex]m_p=9.1\times 10^{-31}\ kg[/tex]
Wavelength, [tex]\lambda_p=4.23\times 10^{-12}\ m[/tex]
We need to find the potential difference. The relationship between potential difference and wavelength is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}[/tex]
[tex]V=\dfrac{h^2}{2q_em_e\lambda^2}[/tex]
[tex]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}[/tex]
[tex]V=6.92\times 10^{34}\ V[/tex]
V = 84109.27 volt
Hence, this is the required solution.
