An alpha particle is accelerated through a potential difference of 100

1.	An alpha particle is accelerated through a potential difference of 100 V. Calculate:(i) The speed acquired by the alpha particle, and (ii) The de-Broglie wavelength associated with it. (Take mass of alpha particle = 6.4 x 10-27 kg)
Answer» (i) Kinetic energy of alpha particle K.E. = qv \(\frac12mv^2= 2e \times100\) \(\frac12mv^2= 200 \,ev\) \(v^2 = \frac{400\,ev}{m}\) \(v= \sqrt{\frac{400\,ev}{m}}\) \(v = \sqrt{\frac{400\times1.6\times10^{19}}{6.4\times10^{27}}}\) \(v =\sqrt{\frac{640}{64}\times10^8}\) \(v = \sqrt{100\times10^8}\) v = 10 x 10⁴ m/s (ii) Kinetic energy gained by alpha particle by potential difference 100 V is \(\frac{p^2}{2m_\alpha}= q_\alpha v\) \(\frac{p^2}{2m_\alpha}= 2\,ev\) \(p^2 = 4m_\alpha ev\) \(p = \sqrt{4m_\alpha ev}\) \(p= \sqrt{4\times6.4\times10^{-27}\times1.6\times10^{-19}\times100}\) \(p = \sqrt{4096\times10^{-46}}\) \(p = 64 \times 10^{-23}\) de-Broglie wavelength, \(\lambda=\frac hp\) \(\lambda=\frac{6.64\times10^{-34}}{64\times10^{-23}}\) \(\lambda = 0.103 \times 10^{-11}\) \(\lambda = 0.0103 \times 10^{-10}\) \(\lambda = 0.0103 \,A^o\)

An alpha particle is accelerated through a potential difference of 100 V. Calculate:(i) The speed acquired by the alpha particle, and (ii) The de-Broglie wavelength associated with it. (Take mass of alpha particle = 6.4 x 10-27 kg)

Answer»

(i) Kinetic energy of alpha particle K.E. = qv

\(\frac12mv^2= 2e \times100\)

\(\frac12mv^2= 200 \,ev\)

\(v^2 = \frac{400\,ev}{m}\)

\(v= \sqrt{\frac{400\,ev}{m}}\)

\(v = \sqrt{\frac{400\times1.6\times10^{19}}{6.4\times10^{27}}}\)

\(v =\sqrt{\frac{640}{64}\times10^8}\)

\(v = \sqrt{100\times10^8}\)

v = 10 x 10⁴ m/s

(ii) Kinetic energy gained by alpha particle by potential difference 100 V is

\(\frac{p^2}{2m_\alpha}= q_\alpha v\)

\(\frac{p^2}{2m_\alpha}= 2\,ev\)

\(p^2 = 4m_\alpha ev\)

\(p = \sqrt{4m_\alpha ev}\)

\(p= \sqrt{4\times6.4\times10^{-27}\times1.6\times10^{-19}\times100}\)

\(p = \sqrt{4096\times10^{-46}}\)

\(p = 64 \times 10^{-23}\)

de-Broglie wavelength,

\(\lambda=\frac hp\)

\(\lambda=\frac{6.64\times10^{-34}}{64\times10^{-23}}\)

\(\lambda = 0.103 \times 10^{-11}\)

\(\lambda = 0.0103 \times 10^{-10}\)

\(\lambda = 0.0103 \,A^o\)