An \alpha-particle has a mass approximately 4 times that of a proton and carries 2 times t… — JAMB UTME

An $\alpha$ -particle has a mass approximately $4$ times that of a proton and carries $2$ times the elementary charge of a proton. If both particles are accelerated from rest through the same potential difference $V$ , what is the ratio of the de Broglie wavelength of the $\alpha$ -particle to that of the proton?

$1 : 2\sqrt{2}$ Answer
B
$1 : 8$
C
$2\sqrt{2} : 1$
D
$1 : 2$

Answer

The ratio of the de Broglie wavelength of the

\alpha

-particle to that of the proton is

1 : 2\sqrt{2}

The de Broglie wavelength of a particle accelerated by a potential difference

V

is given by

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

. Taking the ratio of wavelengths yields

\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}

Step-by-Step Solution

Relate de Broglie wavelength to accelerating potential

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2mqV}}

The kinetic energy gained by a charged particle accelerated through a potential difference

V

E_k = qV

Express wavelengths for both particles

\lambda_p = \frac{h}{\sqrt{2 m_p e V}}

and

\lambda_\alpha = \frac{h}{\sqrt{2 m_\alpha q_\alpha V}}

Setting up the parametric expressions for proton (

m_p, e

) and

\alpha

-particle (

m_\alpha, q_\alpha

Substitute relative values

m_\alpha = 4m_p

and

q_\alpha = 2e

\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16 m_p e V}} = \frac{h}{4\sqrt{m_p e V}}

Using the given relationships between mass and charge of the two particles.

Calculate the ratio

\frac{\lambda_\alpha}{\lambda_p}

\frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{4\sqrt{m_p e V}}}{\frac{h}{\sqrt{2 m_p e V}}} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}

Simplifying the fractional ratio of wavelengths.

Key Concept

de Broglie Wavelength of Charged Particles Accelerated in Electric Fields

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