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Chapter 19: Problem 28

If proton and \(\alpha\)-particles are accelerated by the same potential difference, then their De-Broglie wavelength will be in the ratio of (A) \(\sqrt{2}\) (B) 2 (C) \(2 \sqrt{2}\) (D) 4

Short Answer

Expert verified
The ratio of the De-Broglie wavelength of the proton to the \(\alpha\)-particle is 4 (Option D).

Step by step solution

01

De-Broglie wavelength formula

Recall the De-Broglie wavelength formula, which relates the wavelength (位) of a particle to its momentum (p) as follows: \[\lambda = \frac{h}{p}\] where \(h\) is the Planck's constant.
02

Relation between Kinetic Energy and Momentum

To find the relationship between the Kinetic Energy (KE) of a particle and its momentum (p), we can use the following formula: \[p = \(\sqrt{2m(K_{E})}\] where \(m\) is the mass of the particle.
03

Kinetic energy gained from potential difference

When a charged particle is accelerated through a potential difference (V), it gains kinetic energy (KE) given by: \[K_{E} = qV\] where \(q\) is the charge of the particle.
04

Find the De-Broglie wavelengths for the proton and \(\alpha\)-particle

Using the expressions from Steps 2 and 3, we can find the De-Broglie wavelengths (位) for the proton (p) and \(\alpha\)-particle (伪) separately: For the proton: \[位_{p} = \frac{h}{\sqrt{2m_{p}K_{E_{p}}}} = \frac{h}{\sqrt{2m_{p}(eV)}}\] For the \(\alpha\)-particle: \[位_{伪} = \frac{h}{\sqrt{2m_{伪}K_{E_{伪}}}} = \frac{h}{\sqrt{2m_{伪}(2eV)}}\]
05

Find the ratio of their De-Broglie wavelengths

Find the ratio of the De-Broglie wavelengths of the proton to the \(\alpha\)-particle: \[\frac{位_{p}}{位_{伪}} = \frac{\frac{h}{\sqrt{2m_{p}(eV)}}}{\frac{h}{\sqrt{2m_{伪}(2eV)}}}\] Simplify the expression: \[\frac{位_{p}}{位_{伪}} = \frac{\sqrt{2m_{伪}(2eV)}}{\sqrt{2m_{p}(eV)}} = \sqrt{\frac{2m_{伪}(2eV)}{m_{p}(eV)}} = \sqrt{\frac{4m_{伪}}{m_{p}}}\] Since the mass of an \(\alpha\)-particle is 4 times the mass of a proton (\(m_{伪} = 4m_{p}\)), we get: \[\frac{位_{p}}{位_{伪}} = \sqrt{\frac{4(4m_{p})}{m_{p}}} = \sqrt{16} = 4\]
06

Conclusion

So, the ratio of the De-Broglie wavelength of the proton to the \(\alpha\)-particle is 4 (Option D).

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