The kinetic energy of an electron, an -particle, and a proton are… - Solution | JEE Physics

MCQMediumJEE 2023de Broglie Relation

The kinetic energy of an electron, an $\alpha$ -particle, and a proton are given as $4K$ , $2K$ , and $K$ respectively. The de-Broglie wavelength associated with electron ( $\lambda_e$ ), $\alpha$ -particle ( $\lambda_{\alpha}$ ), and the proton ( $\lambda_p$ ) are as follows:

A
$\lambda_{\alpha} > \lambda_p > \lambda_e$
B
$\lambda_{\alpha} = \lambda_p > \lambda_e$
C
$\lambda_{\alpha} = \lambda_p < \lambda_e$
D
$\lambda_{\alpha} < \lambda_p < \lambda_e$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The kinetic energies are $4K$ for the electron, $2K$ for the $\alpha$ -particle, and $K$ for the proton.

Find: The correct order of de-Broglie wavelengths $\lambda_e$ , $\lambda_{\alpha}$ , and $\lambda_p$ .

Use the de-Broglie relation and the kinetic energy relation:

\lambda = \frac{h}{p}

and

K = \frac{p^2}{2m}

So,

p = \sqrt{2mK}

Therefore,

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

Comparison Using Mass and Kinetic Energy

For the electron:

\lambda_e = \frac{h}{\sqrt{2m_e(4K)}} = \frac{h}{\sqrt{8m_eK}}

For the proton:

\lambda_p = \frac{h}{\sqrt{2m_pK}}

For the $\alpha$ -particle, its mass is approximately $4m_p$ . Hence,

\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2K)}} = \frac{h}{\sqrt{16m_pK}} = \frac{h}{4\sqrt{m_pK}}

Since the electron mass is much smaller than the proton mass, the denominator for $\lambda_e$ is the smallest among the three, so $\lambda_e$ is the largest. Also,

\sqrt{16m_pK} > \sqrt{2m_pK}

therefore,

\lambda_{\alpha} < \lambda_p

Thus,

\lambda_e > \lambda_p > \lambda_{\alpha}

Therefore, the correct option is D.

Common mistakes

Using only kinetic energy to compare wavelengths and ignoring the particle masses is incorrect, because $\lambda = \frac{h}{\sqrt{2mK}}$ depends on both $m$ and $K$ . Compare the product $mK$ for each particle.
Assuming the $\alpha$ -particle and proton have the same mass is wrong. An $\alpha$ -particle has approximately four times the proton mass, so its wavelength becomes smaller for comparable kinetic energy.
Reversing the inverse relation between wavelength and momentum is a common error. Since $\lambda = \frac{h}{p}$ , a larger momentum means a smaller wavelength, not a larger one.

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Answer

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