Question 358: Dual Nature of Matter and Radiation

Question

Kinetic energy of a proton is equal to energy '$$E$$' of a photon. Let '$$\lambda_1$$' be the de-Broglie wavelength of proton and '$$\lambda_2$$' is the wavelength of photon. If $$\frac{\lambda_1}{\lambda_2} \alpha E^n$$, then the value of '$$n$$' is

Accepted Answer

To find the value of '$$n$$', we need to analyze the expressions for the de-Broglie wavelength of the proton and the wavelength of the photon in terms of the energy '$$E$$'.
The kinetic energy of the proton is given by:
$$E_{	ext{proton}} = \frac{1}{2}mv^2$$
where $$m$$ is the mass of the proton and $$v$$ is its velocity.
The de-Broglie wavelength $$\lambda_1$$ of the proton is:
$$\lambda_1 = \frac{h}{mv}$$
Using the expression for kinetic energy, we can write:
$$E = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E}{m}}$$
Substituting this into the expression for the de-Broglie wavelength:
$$\lambda_1 = \frac{h}{m\sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2mE}}$$
Next, consider the energy of the photon:
$$E_{	ext{photon}} = E = \frac{hc}{\lambda_2}$$
Solving for the wavelength $$\lambda_2$$ of the photon:
$$\lambda_2 = \frac{hc}{E}$$
Now we need to find the ratio $$\frac{\lambda_1}{\lambda_2}$$:
$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} = \frac{hE}{hc\sqrt{2mE}} = \frac{E}{c\sqrt{2mE}} = \frac{E^{\frac{1}{2}}}{c\sqrt{2m}}$$
From this expression, it is clear that:
$$\frac{\lambda_1}{\lambda_2} = \alpha E^{\frac{1}{2}}$$
where $$\alpha$$ is a constant involving other parameters like Planck's constant and mass of the proton.
Therefore, the value of '$$n$$' is:
$$\boxed{\frac{1}{2}}$$

Answer

$$\frac{1}{2}$$

Answer

$$\frac{1}{4}$$

Answer

2

Answer

4

Dual Nature of Matter and Radiation

Kinetic energy of a proton is equal to energy '$E' of a photon. Let '\lambda_1' be the de-Broglie wavelength of proton and '\lambda_2' is the wavelength of photon. If \frac{\lambda_1}{\lambda_2} \alpha E^n, then the value of 'n$' is

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