An electron, an \(\alpha \) particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wave length?
Speed
00:00
05:11
An electron, an \(\alpha \) particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wave length?
Answer
For a particle, de Broglie wavelength,
\(\lambda =\frac h p\)
The relation between wavelength and kinetic energy of a particle is
\(K=\frac{p^2}{2m}\)
\(p=\frac h{\lambda }\)
\(K=\frac{\left(\frac h{\lambda }\right)^2}{2m}\)
\(\lambda =\frac h{\sqrt{2\mathit{mK}}}\)
From the above relation it can be concluded that for the same kinetic energy, de Broglie’s wavelength of a particle is inversely proportional to its mass.
Mass of a proton, \(m_p=1836m_e\)
Mass of an \(\alpha \) particle, \(m_{\alpha }=4m_p\)
Therefore the masses are in the order, \(m_{\alpha }>m_p>m_e\) .
Hence, the \(\alpha \) particle has the shortest de Broglie’s wavelength.
\(\lambda =\frac h p\)
The relation between wavelength and kinetic energy of a particle is
\(K=\frac{p^2}{2m}\)
\(p=\frac h{\lambda }\)
\(K=\frac{\left(\frac h{\lambda }\right)^2}{2m}\)
\(\lambda =\frac h{\sqrt{2\mathit{mK}}}\)
From the above relation it can be concluded that for the same kinetic energy, de Broglie’s wavelength of a particle is inversely proportional to its mass.
Mass of a proton, \(m_p=1836m_e\)
Mass of an \(\alpha \) particle, \(m_{\alpha }=4m_p\)
Therefore the masses are in the order, \(m_{\alpha }>m_p>m_e\) .
Hence, the \(\alpha \) particle has the shortest de Broglie’s wavelength.
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
Correct48
Incorrect0
Still Have Question?
Watch More Related Solutions
Load More
More Solution Recommended For You