The kinetic energy of an electron, \alpha -particle and a proton are given as 4 ~K, 2 ~K and K respectively. The de-Broglie wavelength associated with electron (\lambda e), \alpha -particle ((\lambda \alpha) and the proton (\lambda p) are as follows:

\lambda \alpha\lt \lambda p\lt \lambda e

\lambda \alpha\gt \lambda p\gt \lambda e

\lambda \alpha=\lambda p\lt \lambda e

\lambda \alpha=\lambda p\gt \lambda e

Dual Nature of Radiation PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q51 with Solution

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Correct answer: A

For a non-relativistic particle, de-Broglie wavelength is

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mT}}

where $m$ is mass and $T$ is kinetic energy.

So,

\lambda \propto \frac{1}{\sqrt{mT}}

Given kinetic energies:

electron: $T_e = 4K$
$\alpha$ -particle: $T_\alpha = 2K$
proton: $T_p = K$

Masses:

electron: $m_e$
proton: $m_p$
$\alpha$ -particle: $m_\alpha \approx 4m_p$

Compare $\alpha$ -particle and proton:

\lambda_\alpha \propto \frac{1}{\sqrt{m_\alpha T_\alpha}} = \frac{1}{\sqrt{(4m_p)(2K)}} = \frac{1}{\sqrt{8m_pK}}

\lambda_p \propto \frac{1}{\sqrt{m_pK}}

Thus,

\lambda_\alpha = \frac{\lambda_p}{\sqrt{8}} < \lambda_p

Compare proton and electron:

\lambda_e \propto \frac{1}{\sqrt{m_e(4K)}} = \frac{1}{2\sqrt{m_eK}}

\lambda_p \propto \frac{1}{\sqrt{m_pK}}

So,

\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_pK}}{2\sqrt{m_eK}} = \frac{1}{2}\sqrt{\frac{m_p}{m_e}}

Since

\frac{m_p}{m_e} \approx 1836

we get

\frac{\lambda_e}{\lambda_p} \approx \frac{1}{2}\sqrt{1836} > 1

Hence,

\lambda_e > \lambda_p

Therefore the order is

\lambda_\alpha < \lambda_p < \lambda_e

So the correct option is A.

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View written solutionFree

Correct answer: A

For a non-relativistic particle, de-Broglie wavelength is

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mT}}

where $m$ is mass and $T$ is kinetic energy.

So,

\lambda \propto \frac{1}{\sqrt{mT}}

Given kinetic energies:

electron: $T_e = 4K$
$\alpha$ -particle: $T_\alpha = 2K$
proton: $T_p = K$

Masses:

electron: $m_e$
proton: $m_p$
$\alpha$ -particle: $m_\alpha \approx 4m_p$

Compare $\alpha$ -particle and proton:

\lambda_\alpha \propto \frac{1}{\sqrt{m_\alpha T_\alpha}} = \frac{1}{\sqrt{(4m_p)(2K)}} = \frac{1}{\sqrt{8m_pK}}

\lambda_p \propto \frac{1}{\sqrt{m_pK}}

Thus,

\lambda_\alpha = \frac{\lambda_p}{\sqrt{8}} < \lambda_p

Compare proton and electron:

\lambda_e \propto \frac{1}{\sqrt{m_e(4K)}} = \frac{1}{2\sqrt{m_eK}}

\lambda_p \propto \frac{1}{\sqrt{m_pK}}

So,

\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_pK}}{2\sqrt{m_eK}} = \frac{1}{2}\sqrt{\frac{m_p}{m_e}}

Since

\frac{m_p}{m_e} \approx 1836

we get

\frac{\lambda_e}{\lambda_p} \approx \frac{1}{2}\sqrt{1836} > 1

Hence,

\lambda_e > \lambda_p

Therefore the order is

\lambda_\alpha < \lambda_p < \lambda_e

So the correct option is A.

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