Question

A proton is accelerated through a potential difference of 400V to have same de broglie wavelength what potential difference must be applied across doubly ionised 8O16 atom
A.50V. B.12.5V
C.100V. D.none of these

Answer 1

Answer:

In order to have same de-broglie wavelength we must have to apply 12.5 V potential difference

Explanation:

As we know that the de Broglie wavelength of a particle is given as

$\lambda = \frac{h}{mv}$

so for proton we will have

$KE = qV$

so momentum of photon is given as

$P = \sqrt{2mK}$

$P = \sqrt{2mqV}$

so we have

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

now similarly for oxygen we will have

mass of oxygen = $m_{o} = 16 m$

charge on oxygen = $Q = 2q$

so we have

$P = \sqrt{2(16m)(2q) V'}$

now since de Broglie wavelength is same for both so we have

$\frac{h}{\sqrt{64 mqV'}} = \frac{h}{\sqrt{2mqV}}$

so we have

$64 V' = 2 V$

$V' = \frac{V}{32}$

$V' = \frac{400}{32}$

$V' = 12.5 Volts$

#Learn

Topic : De-Broglie Wavelength

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