A proton when accelerated through a potential difference of V volt has a wavelength $λ$ associated with it. An $α$ - particle in order to have the same wavelength $λ$ must be accelerated through a p.d. of

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Answer 1

For proton, $λ_{p} = \frac{0 . 2 8 6 \times 1 0 ^{- 10}}{ν} m$
where V is p.d
For $α - p a r t i c l e, λ_{α} = \frac{0 . 1 0 1 \times 1 0 ^{- 10}}{ν} m$

so, keeping same $α$ for proton and $α - p a r t s i d e$

$1 = \frac{λ _{p}}{λ _{α}} = \frac{0 . 2 8 6 \times V _{α}}{0 . 1 0 1 \times V _{P}}$

$0.101 \times V_{P} = 0.286 \times V_{α}$
$(0.101)^{2} \times V_{P} = (0.286)^{2} V_{α}$

$\frac{( 0 . 1 0 1 ) ^{2} ν}{( 0 . 2 8 6 ) ^{2}} = V_{α}$

$(∵ V_{P} = V)$

$\Rightarrow V_{α} = \frac{V}{8} v o l t$

So, the answer is option (A).

Answer 2

For proton,

λ_{p} = \frac{0 . 2 8 6 \times 1 0 ^{- 10}}{ν} m

where V is p.d
For

α - p a r t i c l e, λ_{α} = \frac{0 . 1 0 1 \times 1 0 ^{- 10}}{ν} m

so, keeping same

α

for proton and

α - p a r t s i d e

1 = \frac{λ _{p}}{λ _{α}} = \frac{0 . 2 8 6 \times V _{α}}{0 . 1 0 1 \times V _{P}}

0.101 \times V_{P} = 0.286 \times V_{α}

(0.101)^{2} \times V_{P} = (0.286)^{2} V_{α}

\frac{( 0 . 1 0 1 ) ^{2} ν}{( 0 . 2 8 6 ) ^{2}} = V_{α}

(∵ V_{P} = V)

\Rightarrow V_{α} = \frac{V}{8} v o l t

So, the answer is option (A).