Dual Nature of Radiation and Matter Test - 81

Correct answer: Option (B).

Hint: Apply the formula of de-broglie wavelength in relation to kinetic energy

Step 1: Find out the relation between de -broglie wavelength and kinetic energy.

De – Broglie wavelength is the wavelength that is associated with an object in relation to its momentum and mass.

The de -broglie wavelength can be written as:

$$\lambda=\dfrac{h}{p}$$ (where, $$\lambda$$ is the de – Broglie wavelength, $$h$$ is Planck’s constant and $$p$$ is the momentum) ….(i)

Now, we know that,

$$K.E=\dfrac{p^2}{2m}$$ (where, $$K.E$$ is the kinetic energy)

$$\Rightarrow p^2=2mK.E$$

$$\Rightarrow p=\sqrt{2mK.E}$$

Putting the value of $$p$$ in equation (i), we get,

$$\lambda=\dfrac{h}{\sqrt{2mK.E}}$$ ….(ii)

Step 2: Calculate the ratio of de-broglie wavelength of a proton and an alpha particle of same energy.

According to the given problem, both the proton and the alpha particle has the same energy.

A proton is designated as $$H^1_1$$

An alpha particle is designated as $$He^4_2$$

So, the mass of an alpha particle is four times the mass of a proton.

Let the mass of proton be $$m$$

Mass of alpha particle will be $$4m$$

Since, the energy is same and $$h$$ is constant, equation (ii) can be written as:

$$\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha}}{m_{proton}}}$$

$$\Rightarrow\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=\sqrt{\dfrac{4m}{m}}$$

$$\Rightarrow\dfrac{\lambda_{proton}}{\lambda_{\alpha}}=2$$

Hence, the ratio is $$2$$.

The correct answer is option (B).