A deuteron and α-particle have the same kinetic energy then find the ratio of their de-Broglie wavelength?

Question

Question

A deuteron and α-particle have the same kinetic energy then find the ratio of their de-Broglie wavelength?

1
√2
2√2
2

Answer 1

Concept:

According to the De-Broglie hypothesis, particles behave as waves & these are called matter waves. The wavelength of the particle is called De-Broglie wavelength and it is given as

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

Where m = mass of the charged particle and KE = Kinetic energy

Calculation:

A deuteron and α-particle have the same kinetic energy then KE₁ = KE₂

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

\(\frac{{{\lambda _d}}}{{{\lambda _\alpha }}} = \frac{{\sqrt {{m_\alpha }} }}{{\sqrt {{m_d}} }}\)

= \(\frac{{\sqrt {4{m_p}} }}{{\sqrt {2{m_p}} }} = \sqrt 2\)

Where; m_p is the mass of a proton.

Note:

1). Mass of deuteron is 2 times of proton mass and the Charge of the deuteron is equal to the charge of the proton.

2). Mass of α-particle is 4 times of proton mass and the Charge of the α-particle is equal to the 2 times of charge of the proton.

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A deuteron and α-particle have the same kinetic energy then find the ratio of their de-Broglie wavelength?

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