Question

The accelerating voltage of a proton is increased to twice. How will its de-Broglie wavelength change? Explain.​

Answer 1

Answer:

Hi there!!!

Explanation:

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refer to the picture....

Answer 2

The change in de-Broglie wavelength is 0.414 units.

Given:

The accelerating voltage of a proton is increased to twice.

To Find:

The change in de-Broglie wavelength.

Solution:

To find the change in de-Broglie wavelength, we will follow the following steps:

As we know,

The relation between de-Broglie wavelength and mass and voltage is given by the formula:

$λ = \frac{h}{ \sqrt{2mev} }$

Here,

λ is de Broglie wavelength, m is the mass and e is the charge of an electron.

v is the voltage.

As the wavelength is inversely proportional to the square root of voltage, we can write the change in wavelength as:

$\frac{λ2 -λ1 }{λ1} = \frac{ \sqrt{v1} }{ \sqrt{v2} } - 1 = \frac{ \sqrt{v1} - \sqrt{v2} }{ \sqrt{v1} }$

v1 and v2 are the initial and final voltage respectively.

λ1 and λ2 are the initial and final wavelengths respectively.

v2 = 2v1

Now,

on putting value as get,

Change in wavelength =

$\frac{ \sqrt{v1} - \sqrt{2v1} }{ \sqrt{v1} } = \frac{ \sqrt{1} - \sqrt{2} }{ \sqrt{1} } = \frac{1 - \sqrt{2} }{1} = \frac{1 - 1.414}{1} = - 0.414 \: \: units$

As wavelength can't be negative which means it decreases by doubling the voltage.

So,

Change in wavelength = 0.414 units

Henceforth, The change in de-Broglie wavelength is 0.414 units.

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