A deuteron of mass 2m_p and charge e, and an \alpha-particle of mass 4m_p and charge 2e, a… — JAMB UTME

A deuteron of mass $2m_p$ and charge $e$ , and an $\alpha$ -particle of mass $4m_p$ and charge $2e$ , are accelerated from rest through potential differences of $V_d$ and $V_\alpha$ respectively. If both particles acquire equal de Broglie wavelengths, what is the ratio of their accelerating potential differences, $\frac{V_d}{V_\alpha}$ ?

$4 : 1$ Answer
B
$1 : 4$
C
$2 : 1$
D
$16 : 1$

Answer

The ratio of the accelerating potential differences

\frac{V_d}{V_\alpha}

4 : 1

Using the de Broglie wavelength equation

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

, setting

\lambda_d = \lambda_\alpha

gives

m_d q_d V_d = m_\alpha q_\alpha V_\alpha

. Substituting

m_d = 2m_p

q_d = e

m_\alpha = 4m_p

, and

q_a = 2e

yields

2 V_d = 8 V_\alpha

, which simplifies directly to

\frac{V_d}{V_\alpha} = 4 : 1

Step-by-Step Solution

Express the de Broglie wavelength in terms of particle mass, charge, and accelerating potential difference.

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

Kinetic energy gained by a charged particle in an electric field is

E_k = qV

, and momentum is

p = \sqrt{2mE_k} = \sqrt{2mqV}

Set the de Broglie wavelengths of the deuteron and

\alpha

-particle equal to each other.

\frac{h}{\sqrt{2 m_d q_d V_d}} = \frac{h}{\sqrt{2 m_\alpha q_\alpha V_\alpha}} \implies m_d q_d V_d = m_\alpha q_\alpha V_\alpha

Equal wavelengths mean their denominators in the de Broglie expression must be equal.

Substitute the given values for mass and charge into the equality.

(2m_p)(e) V_d = (4m_p)(2e) V_\alpha \implies 2 m_p e V_d = 8 m_p e V_\alpha

Deuteron has mass

2m_p

and charge

e

;

\alpha

-particle has mass

4m_p

and charge

2e

Solve for the ratio

\frac{V_d}{V_\alpha}

\frac{V_d}{V_\alpha} = \frac{8}{2} = 4

Simplifying the algebraic equation by cancelling common terms

m_p

and

e

Key Concept

De Broglie Wavelength of Accelerated Charged Particles

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