An $α$ - particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

Question

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Answer 1

Given that;

$V_{α} = V_{P r}$

Since:- $λ = \frac{h}{p}$

$\Rightarrow λ = \frac{h}{2 m E}$

Where $E = k i n e t i c E n e r g y = e V$ [V=acceleration potential]

$\Rightarrow λ_{P r} = \frac{h}{2 m _{p} ( 2 e ) V _{P r}}$

Whereas; $λ_{α} = \frac{h}{m _{α} ( 2 e ) V _{α}}$

Where $m_{α} = 4 m_{P r}$

and $V_{α} = V_{P r} = V$

$\Rightarrow λ_{P r} = \frac{h}{2 m _{p} ( e ) V}$ ...........(1)

and $λ_{α} = \frac{h}{2 ( 4 m p ) ( 2 e ) V}$ ...........(2)

From (1) and (2)

$\frac{λ _{P r}}{λ _{α}} = \frac{h}{2 m _{p} ( e ) V} \times \frac{2 ( 4 m _{p} ) ( 2 e ) V}{h}$

$= \frac{2 m _{p} e V ( 8 )}{2 m _{p} e V}$

$= 8 : 1$

Answer 2

Given that;

V_{α} = V_{P r}

Since:-

λ = \frac{h}{p}

\Rightarrow λ = \frac{h}{2 m E}

Where

E = k i n e t i c E n e r g y = e V

[V=acceleration potential]

\Rightarrow λ_{P r} = \frac{h}{2 m _{p} ( 2 e ) V _{P r}}

Whereas;

λ_{α} = \frac{h}{m _{α} ( 2 e ) V _{α}}

Where

m_{α} = 4 m_{P r}

and

V_{α} = V_{P r} = V

\Rightarrow λ_{P r} = \frac{h}{2 m _{p} ( e ) V}

...........(1)

and

λ_{α} = \frac{h}{2 ( 4 m p ) ( 2 e ) V}

...........(2)

From (1) and (2)

\frac{λ _{P r}}{λ _{α}} = \frac{h}{2 m _{p} ( e ) V} \times \frac{2 ( 4 m _{p} ) ( 2 e ) V}{h}

= \frac{2 m _{p} e V ( 8 )}{2 m _{p} e V}

= 8 : 1