Question

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The kinetic energy of a proton and that of an alpha- particle are 4eV and 1eV, respectively. The ratio of the de-Brogile wavelengths associated with them will be

(a) 2:1
(b) 1:1
(c) 1:2
(d) 4:1

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Answer 1

Answer:

ratio of de broglie wavelength of proton and alpha particle is given as

$\frac{\lambda_p}{\lambda_a} = 1 : 1$

Explanation:

As we know that the de Broglie wavelength is given as

$\lambda = \frac{h}{mv}$

here we know that the momentum and kinetic energy is related as

$mv = \sqrt{2mK}$

now we have

$\lambda = \frac{h}{\sqrt{2mK}}$

now ratio of de broglie wavelength of proton and alpha particle is given as

$\frac{\lambda_p}{\lambda_a} = \frac{\sqrt{m_aK_a}}{\sqrt{m_pK_p}}$

here we know that

$m_p = m$

$K_p = 4 eV$

$m_a = 4 m_p$

$K_a = 1 eV$

now we have

$\frac{\lambda_p}{\lambda_a} = \frac{\sqrt{4m(1 eV)}}{\sqrt{m(4 eV)}}$

$\frac{\lambda_p}{\lambda_a} = 1 : 1$

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Topic : De Broglie wavelength

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Answer 2

Answer:

b) 1:1

Explanation:

The explanation is below in the picture.

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