An electron and an alpha particle are accelerated by the same potential difference. Let λe and λα denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then: (a) λe > λα (b) λe = 4λα (c) λe = λα (d) λe < λα

Answer: (a). For a charge q accelerated through V: λ = (h)/(√(2mqV)), so λ ∝ (1)/(√(mq)) for the same V. (λe)/(λα) = √((mα qα)/(me qe)) = √((4mp · 2e)/(me · e)) = √((8mp)/(me)) 1 So the answer is (a) λe > λα.

An electron and an alpha particle are accelerated by the same… | NEET 2024

Q53NEET 2024Easy

Matter Waves: de Broglie Wavelength

An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_\alpha$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

Show answer and solution

Answer: (a)

For a charge $q$ accelerated through $V$ : $\lambda = \dfrac{h}{\sqrt{2mqV}}$ , so $\lambda \propto \dfrac{1}{\sqrt{mq}}$ for the same $V$ .
$\dfrac{\lambda_e}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_e q_e}} = \sqrt{\dfrac{4m_p \cdot 2e}{m_e \cdot e}} = \sqrt{\dfrac{8m_p}{m_e}} \gg 1$
So the answer is (a) $\lambda_e > \lambda_\alpha$ .

Report a mistake

Matter Waves: de Broglie Wavelength question from NEET 2024

More questions on Matter Waves: de Broglie Wavelength