Question

An alpha particle and a proton are accelerated
separately by the potential difference of 2 V and ZV
respectively but ratio of de Broglie wavelength of alpha
particle to proton is 2. The value of 'Z' is​

Answer 1

Given :

An alpha particle and a proton are accelerated  separately by the potential difference of 2 V and Z V.

Ratio of De Broglie wavelength of alpha  particle to proton is 2.

To Find :

The value of 'Z' .

Solution :

De Broglie wavelength of alpha particle of mass $m$ , carrying charge $q$ accelerating through potential V is :

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

Let , mass of proton is m and charge on it is q .

Therefore , mass of alpha particle is 4m and charge is 2q .

So ,

$\lambda_\alpha=\dfrac{h}{\sqrt{2mqV}}$

and $\lambda_p=\dfrac{h}{\sqrt{2(4m)(2q)Z}}$

It is given that :

$\dfrac{\lambda_\alpha}{\lambda_p}=\dfrac{2}{1}$

$\dfrac{\dfrac{h}{\sqrt{2mqV}}}{\dfrac{h}{\sqrt{2(4m)(2q)Z}}}=\dfrac{2}{1}\\\\\\\dfrac{V}{8Z}=\dfrac{4}{1}\\\\Z=\dfrac{V}{32}$

Therefore , value of Z is $\dfrac{V}{32}$ .

Hence , this is the required solution .

Answer 2

Explanation:

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